50 AMC Lectures Chapter 33 Logarithms BASIC KNOWLEDGE 1. Definition If a and x are positive real numbers and a ≠ 1, then y is the logarithm to the base a of x. This can be written as y = log a x (1.1) if and only if x = a y (1.2) A quick way to convert the logarithm to the exponent: Example 1: Rewrite the logarithmic forms as the exponential form: (1): log 5 x = y ⇒ 5 y = x (2): log 5 3 = y ⇒ 5 y = 3 (3): log a 1024 = 10 ⇒ a 10 = 1024. (4): log10 1000 = 3 ⇒ 10 3 = 1000. Notes: (1) The logarithm base 10 is called the common logarithm. log x always refers to log base 10, i.e., log x = log 10 x. (2) The logarithm base e is called the natural logarithm. ln x = log e x. (3) log 10 x is always written as log x. (4) Zero and negative numbers have no logarithm expressions. You will not be able to see any expressions like log a 0 , log a (− 3), or . x 10 log − The graphs of logarithmic functions f (x) = log a x all have x-intercept 1, and are increasing when a > 1 and decreasing when a < 1. 228
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50 AMC Lectures Chapter 33 Logarithms
BASIC KNOWLEDGE 1. Definition If a and x are positive real numbers and a ≠ 1, then y is the logarithm to the base a of x. This can be written as
y = loga x (1.1)
if and only if x = a y (1.2) A quick way to convert the logarithm to the exponent:
Example 1: Rewrite the logarithmic forms as the exponential form: (1): log5 x = y ⇒ 5y = x (2): log53 = y ⇒ 5y = 3
(3): loga1024 = 10 ⇒ a10 = 1024. (4): log10 1000 = 3 ⇒ 103 = 1000. Notes: (1) The logarithm base 10 is called the common logarithm. log x always refers to log base 10, i.e., log x = log10 x. (2) The logarithm base e is called the natural logarithm. ln x = loge x. (3) log10 x is always written as log x. (4) Zero and negative numbers have no logarithm expressions. You will not be able to see any expressions like log a 0 , loga (− 3), or . x10log−
The graphs of logarithmic functions f (x) = log a x all have x-intercept 1, and are increasing when a > 1 and decreasing when a < 1.
228
50 AMC Lectures Chapter 33 Logarithms
y = loga x (a > 1) y = loga x (a < 1)
Example 2: (1970 AMC) If and 225log8=a ,15log2=b then
(A) 2ba = (B)
32ba = (C) a = b (D)
2ab = (D)
23ba =
Solution: (B). The exponential form of the given equations is 8a = 225 and 2b = 15.
Since 8 = 23 and 152 = 225, we have (23)a = 225 = 22b, so 3a = 2b and 32ba = .
Example 3: (1974 AMC) If p=3log8 and ,5log3 q= then, in terms of p and q, equals
5log10
(A) pq (B) 5
3 qp + (C) qppq
++ 31 (D)
pqpq
+13
Solution: (D). The exponential form of the given equations is 3 = 8p = 23p and 5 = 3p. 5 = (23p)q = 23pq. We are asked to find i.e. x such that 10x = 5. Since 5 = 23pq, we have 10x = 23pq.
,5log10 x=
10x = 2x · 5x = 2x · 23pqx = 2x(1+3pq). It follows that
x(1 + 3pq) = 3pq and x = pq
pq+13 .
2. Properties of Logarithms: If a, b, x and y are positive real numbers, a ≠ 1, b ≠ 1, and r is any real number, we have
229
50 AMC Lectures Chapter 33 Logarithms
230
0
1
xr
1log =a (2.1)
log =aa (2.2)
a xa =log (2.3) ar
a =log (2.4)
loga b log b c = log a c (2.5) 3. The Laws of Logarithms: Law 1: the Product Identity
loga xy = log a x + log a y (3.1) Proof : Let m = loga x and n = loga y. The exponential form of the equations is am = x and an = y. Multiplying these two equations, we get aman = xy ⇒ am +n = xy Transforming the exponential form to the logarithmic form, we get loga xy = m + n.
Since m = loga x and n = loga y, loga xy = log a x + log a y. Example 4: Calculate:
2log2log3log)3(log 6662
6 +⋅+
Solution: 1. 2log2log3log)3(log 666
26 +⋅+ 2log)2log3(log3log 6666 ++=
2log3log2log)23(log3log 66666 +=+×= .1)23(log6 =×= Law 2: the Quotient Identity
x = loga x − loga y (3.2) loga (
y)
Proof: Let m = loga x and n = loga y. The exponential form of the equations is am = x and an = y Dividing these two equations, we get am / an = x/y ⇒ am − n = x/y
50 AMC Lectures Chapter 33 Logarithms
Use the definition of the logarithm, we get loga (xy
) = m − n.
Since m = loga x and n = loga y, loga (xy
) = loga x − loga y.
Example 5: (1967 AMC) If x is real and positive and grows beyond all bounds, then
approaches: )12(log)56(log 33 +−− xx(A) 0 (B) 1 (C) 3 (D) 4 (E) no finete nimber Solution: (B). For real x > 1, 6x – 5 and 2x + 1 are positive, so both logarithms are defined. Now
)12(log)56(log 33 +−− xx )12
83(log12
836log1256log 333 +
−=+
−+=
+−
=xx
xxx approaches
as x increases beyond all bounds, because 1log3 =12
8+x
then approaches 0.
Law 3: the Power Identity Formula 1: loga xr
= r loga x (3.3) Proof: Since , x rx
ar
aaax )(loglog log=aax log= →loga x r = (r loga x)loga a.
Recall that . Substituting this value into the equation, we have
The domain is the same (0, +∞) for ,log xy = or xxy
2
log= .
For x ∈ (0, +∞), xxx
=2
. Therefore, xy log= and xxy
2
log= are the same.
232
Formula 2: bmnb a
nam loglog =
c
(a, b > 0, a ≠ 1) (3.4)
Proof: Let . bn
am =log
The exponential form of the equations is ncm ba =)( .
Taking the logarithm of both sides yields or bnac am
a loglog = bncm alog= .
bmnc alog= ⇒ b
mnb a
nam loglog = .
Example 7: Calculate )2log4log8)(log5log25log125(log 525125842 ++++ . Solution: 13.
We use the formula bmnb a
nam loglog to simplify the given equation: =
)2log2log2)(log5log315log5log3( 555222 ++++
.135log5log132log35log
313
2
252 ==⋅=
4. Change-of-Base Theorem:
Formula 1: loga x =logb xlogb a
(4.1)
Proof: Let y = loga x. The exponential form of this equation is ay = x. Taking the logarithm on both sides, we get logb ay = logb x ⇒ ylogb a = logb x
Dividing both sides by logb b, we obtain y =logb xlogb a
⇒ loga x =logb xlogb a
.
50 AMC Lectures Chapter 33 Logarithms
Example 8: (1974 AMC) If p=3log8 and ,5log3 q= then, in terms of p and q, equals
5log10
(A) pq (B) 5
3 qp + (C) qppq
++ 31 (D)
pqpq
+13 (E) p2 + q2
Solution: (D). Since we need to find a logarithm in base 10, we must first convert all the given information into that base. This can be done by using (4.1).
From 8log3log3log
10
108 ==p , and
315log5log
10
103 og
q == , we obtain ,8log3log 1010 p=
thus ,3log5log 1010 q= 3
101010 )5
10(log8log5log pqpq == ).5log1( 103 −= pq
Soving this for gives choice (D) as the answer. 5log10
Formula 2: m ≠ 0 (4.2) ,loglog m
aa NN m= Proof:
.logloglog
loglog
logloglog m
amb
mb
b
b
b
ba N
aN
amNm
aNN m====
Formula 3: ,log
1loga
NN
a = N ≠ 1 (4.3)
Proof:
aaNN
NN
Na log
1logloglog == .
Example 9: (1974 AMC). If p=3log8 and ,5log3 q= then, in terms of p and q, equals
5log10
(A) pq (B) 5
3 qp + (C) qppq
++ 31 (D)
pqpq
+13 (E) p2 + q2
Solution: (D).
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50 AMC Lectures Chapter 33 Logarithms
We know that a
bb
a log1log = and can write ,
2log31
2log1
8log13log
33
338 ====p
so ,312log3 p
= and .31
3
315log2log10log
5log5log333
310 pq
pq
qp
qq+
=+
=+
==
234
Formula 4: N
N aa1loglog −= (4.4)
Proof:
NNN aaa
1log)1(loglog 1 −== −
Formula 5: NN
aa 1loglog −= (4.5)
Proof:
NNNa
aa 11 logloglog 1 −== −
−
Formula 6: m ≠ 0 (4.6) ,loglog NmN maa = Proof:
NmNN mm am
aa logloglog == Example 10: (2009 ARML) Compute all real values of x such that
. )(loglog)(loglog 4422 xx = Solution: By the formula (4.6), we have =)(loglog 22 x 2
4444 )log2(log)log2(log2 xx = .
Therefore = ⇒ ⇒
⇒
244 )log2(log x
0log) 42 =− x
)(loglog 44 x
(log4
xx 42
4 log)log2( =
0)(log4 4 x 1log4)( 4 =−xx
We have ⇒ (extraneous). 0log4 =x 140 ==x
Or ⇒ 01log4 2 =−x41log4 =x ⇒ 24 4/1 ==x .
50 AMC Lectures Chapter 33 Logarithms
235
Formula 7: NM
NM
b
b
a
a
loglog
loglog
= (a > 0, a ≠ 1, b > 0, b ≠ 1, M, N > 0, N ≠ 1) (4.7)
Proof :
By the Change-of-Base Theorem, we have MNM
Na
a logloglog
= , and MNM
Nb
b logloglog
= .
Therefore NM
NM
b
b
a
a
loglog
loglog
= .
Nn
N an
a log1log =Formula 8: , N is integer greater than 1. (4.8)
Example 11: Find if 8log 3 if a=3log12 .
Solution: 3(1− a
a)
.
)3log12(log33
12log34log364log8log 333333 −====
= a
aa
)1(3)11(3)13log
1(312
−=−=− .
5. Solving Logarithm Equations If x, y and a ≠ 1 are positive real numbers, x = y if and only if loga x = loga y. Similarly xa = ya if and only if x = y. When we use the laws of logarithms, we need to be careful about the following restrictions.
When we apply Law 1 the Product Identity loga xy = log a x + log a y or Law 2: the
Quotient Identity lx
oga (y
) = loga x − loga y, we need to know that both x and y in the
expressions loga xy and loga (xy
), can be negative. However, in the expressions log a x +
log a y or loga x − loga y, neither of them can be negative.
50 AMC Lectures Chapter 33 Logarithms
Example 12: (2001 ARML Tiebreak #3) Given 21
)log(
)log(=
yxxy , increasing y by 50%
decreases x by a factor of k. Computer k. Solution: Method 1 (Official solution):
Step 1:yx
yxxy log)log(
21)log( == (1)
Step 2: yxxy = (2)
Step 3: 23
1
yx = (3)
Step 4: 31y
x = (4)
Increasing by 50% gives xyy 27
81278
)23(
13
3=⎟⎟
⎠
⎞⎜⎜⎝
⎛⋅= . Thus
278
=k .
Note: The given equation is valid for x, y < 0. From step 2 to step 3 in the official
solution, they separated yx to
yx , which is incorrect for the case x, y < 0.
The correct way should be: Squaring both sides of (2): yxxy =2)( ⇒ 3
1y
=x .
Method 2 (Solution given by Intermediate Algebra from Art of Problem Solving): Multiplying both sides by 2 log (x/y), the equation becomes
yxxy log)log(2 = (1)
So (2) yxyx logloglog2log2 −=+Rearranging this gives 0log3log =+ yx (3) Applying logarithm identities to the left sides gives
236
50 AMC Lectures Chapter 33 Logarithms
)log(logloglog3log 33 xyyxyx =+=+ , so , which mean . Hence if y increases by 50%, meaning that y is multiplied by 3/2, then x must be multiplied by a factor of (2/3)3 = 8/27.
0)log( 3 =xy 13 =xy
Note: The given equation is valid for x, y < 0. From equation (1) to equation (2), we think that the author made a mistake by applying the product identity. We cannot apply the product identity because (2) is not valid when x, y < 0. When we apply Law 3: the Power Identity loga xr
= r loga x, we need to be sure that in
the expressions loga xr, x can be negative if r is even. However, x must be positive in the
expression r loga x. Example 13: (2001 ARML) Compute the largest real value of b such that the solutions to the following equation are integers: . 4
222
2 1010 log)(log xx b =
Solution: Method 1 (from the book "Intermediate Algebra" by Art of Problem Solving): We remove the exponents from the arguments, x2b and x4:
xxb 1010 22
2 log4)log2( = (1)
Noting that xxx 210/1
22 log101loglog 10 == , our expression becomes:
xxb2
22 log
52)log
5( = (2)
Multiplying both sides of (2) by 25 gives us: xxb 2
22
2 log10)(log = ⇒ . 0
2b
log)10log( 222 =− xxb
We have or . 0log2 =x 010log22 =−xb
The first equation gives us x = 1 no matter what b is. The second equation gives us
. /102x =In order for x to be an integer, the expression 10/b2 must be a positive integer. The largest value is 10=b .
237
50 AMC Lectures Chapter 33 Logarithms
Note: We believe that equation (1) is incorrect, since you cannot apply the power identity. The equation is true for x < 0, however (1) will not be
true, so (1) is not equivalent to the given equation.
42
222 1010 log)(log xx b =
Method 2 (Official solution):
xx b1010 2
222
log4)(log = (1)
⇒ (2) 0log4)(log4 1010 22
22 =− xxb
0)1)(log)((log 1010 22
2 =−xbx
238
0
0
If , then x = 1 . log 102=x
If , then 1log 1022 =−xb 22
1log 10 bx = , giving 22
10110 2)2( bbx == .
There are several values that b can hold so that x an integer. If 1±=b , the . If 102=x
310
±=b , the . But the largest value of b such that the second solution an integer
is
92=x
10=b , which gives us x = 2. Note: We believe that both equations (1) and (2) are incorrect. The given equation is valid whether or not x > 0 or x < 0 but (2) and (3) will not be true if x < 0, so (2) and (3) are not equivalent to the given equation. Method 3 (our solution): The correct way should be the following:
42
222 1010 log)(log xx b = ⇒ (We should make sure
that the argument is still positive even when x < 0).
22
222 1010 log2)log( xxb =
Taking b2 out, the equation becomes ⇒ 2
222
22
1010 log2)(log xxb = 0log2)(log 22
222
21010 =− xxb
⇒ . 0log)2log( 22
22
21010 =− xxb
This gives us or . 0log 2210 =x 02log 2
22
10 =−xb
50 AMC Lectures Chapter 33 Logarithms
The first equation results in . Solving for x, we get x = 1 or x = − 1. We can see that both values of x are the solutions to the given equation. Under these values of x, b can be any value. So there is no largest value of b possible.
12 =x
Solving the second equation, we get . 2log 22
210 =xb
Since we want the largest value of b and we can assume that b ≠ 0, we have
22
2
2log 10 bx = ⇒ 22
202102 2)2( bbx ==
Therefore 2210
2120
1 2)2( bbx == or 2210
2120
2 2)2( bbx −=−= . In order for x to be an integer, the expression 10/b2 must be a positive integer. Since b is real number, there are many values that b can hold such that x is an integer for example,
, 1±=b 2±=b , 5±=b , n
b 10±= , where n can be any positive integer. The largest
such value is 10=b . Example 14: Solve )12(log2log 2
11 −+= −− xx xxx
Solution: By Law 3: the Power Identity, loga xr
= r loga x, the given equation can be written as )12(log2log 2
11 −+= −− xxx
xx
1
(1)
So )12(2 2 −+= xxx
2 =x ⇒ x = 1 or x = – 1. x = 1 is extraneous. We obtained an extraneous solution because we missed something when using the law 3. Note that law 3 has three restrictions: 1 – x > 0, 1 – x ≠ 1, and or x < 1. With these restrictions, we can rule out the extraneous solution x = 1.
012 2 >−+ xx
Example 15: Find the value of y if
⎪⎩
⎪⎨⎧
=+−+
=−+
(2) .021log)2(log2
(1) ,0)22(log
55
25
yx
xx.
239
50 AMC Lectures Chapter 33 Logarithms
Solution: Simplifying (1), we get x2 + 2x – 2 = 1. Solving this quadratic for x, we get x = 1, x = – 3. x = – 3 is extraneous, since it yields a negative value that we must take the logarithm of when we substitute x into equation (2).
Simplifying (2), we get 1)2(5 2
=+
yx .
Substituting x = 1 into the above equation yields 59)21(5 2 =+=y . 6. Logarithm Applications Example 16: Find log616 if log1227 = a.
Solution: 4(3 − a)
3+ a.
Method 1:
log616 = )312(log
312log4
36log21
4log26log
16log
12
12
12
12
12
12
×==
a
a
311
)311(4
27log311
)27log311(4
3log1)3log1(4
12
12
12
12
+
−=
+
−=
+−
= .3
)3(4aa
+−
=
Method 2:
3log14
6log2log42log416log
22
266 +
=== (1)
a=+
=×
=3log2
3log3)32(log
3log27log2
2
2
2
3
212 ⇒
aa
−=
323log2 .
aa
aa +
−=
−+
=3
)3(4
321
416log6 .
Example 17: Find (2x)x if 2444 1 =− −xx . (A) .55 (B) 25. (C) .525 (D) 125.
240
50 AMC Lectures Chapter 33 Logarithms
Solution: (C).
2444 1 =− −xx ⇒ 244414 =− xx ⇒ 4x = 32
∴ 22122log42log32log 4
244 +=+=×==x .
2x = 5 ⇒ 5255)2( 212
==+xx .
Example 18: Find the product of α and β if they are two distinct roots of the equation (log 3x)(log 5x) = k.
(A) log3 · log5 – k. (B) log15. (C) .151 (D) 15.
Solution: 151
=αβ .
Since α and β are two distinct roots, we have k=⋅ αα 5log3log (1) k=⋅ ββ 5log3log (2) (1) – (2): ββαα 5log3log5log3log ⋅−⋅ = 0.
Example 19: Find if log xabcd , log mxa = , log nxb = , log pxc = qxd =log . x ≠ 1.
Solution: mnpq
mnp + mpq + mnq + npq.
dcbaabcdx
xxxxxabcd loglogloglog
1log
1log+++
==
241
50 AMC Lectures Chapter 33 Logarithms
xxxx dcba log1
log1
log1
log1
1
+++= .1111
1npqmnqmpqmnp
mnpq
qpnm+++
=+++
=
Example 20: If a, b, c form a geometric sequence and form an arithmetic sequence, find the common difference of the arithmetic sequence.
,log ac ,log cb log ba
Solution: 3/2. Let . dbcda abc −==+ logloglog
dab
bcd
ca
−==+loglog
loglog
loglog
bc
aadb
ccda
loglog
logloglog
logloglog
=−
=+
∴ bc
acadbcda
loglog
loglogloglogloglog
=+
−++ .
(1) loglog
log
loglog
bc
acacdab
=+
Substituting b2 = ac into (1) yields 2loglog2loglog ccacdab ==+ .
Therefore 23
)log(
)log(
log
log
log
log 2322
====
ac
ac
acaca
c
ac
abc
d .
Example 21: Find the maximum value of xx −+ 7loglog if – 1 ≤ x ≤ 2, x ≠ 0.
Solution: 1.
449)
27(log7log7loglog 22 −−=−=−+ xxxxx .
We know that – 1 ≤ x ≤ 2, 23
27
29
−≤−≤− x , 481)
27(
49 2 ≤−≤ x .
8449)
27(10 2 ≤−−≤− x .
242
50 AMC Lectures Chapter 33 Logarithms
∴ 10449)
27( 2 ≤−−x .
Equality occurs when x = 2. Therefore the greatest value of xx −+ 7loglog is log 10 = 1.
Example 22: Find a/b if baba loglog)](21log[2 +=− .
Solution: 223 +=ba
For log a, log b, and log[12
(a − b)] to be defined, the following statement must be true: a
> b > 0.
The given equation can be written as )log()](21log[ 2 abba =− .
∴ abba =− 2)](21[ ⇒ a2 – 6ab + b2 = 0 .
Since that b > 0, we can divide both sides of the equation by b2 and obtain:
0162
=+⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛
ba
ba .
Solving for a/b, we get 223 +=ba or 223 −=
ba . The latter answer is extraneous
since 1223 <− . Example 23: (1983 AIME) Let x, y and z all exceed 1 and let w be a positive number such that and , 40log =wy24log =wx .12log =wxyz Find .log wz
Solution: 60. Method 1: Changing the given logarithms into into exponential forms, we get
, 40 wy =24 wx = , wxyz =12) ( .
51
103
211212
12 www
wyx
wz === .
Therefore and . 60zw = 60log =wz
243
50 AMC Lectures Chapter 33 Logarithms
Method 2:
244
tLet . We then have wz =logt
wz loglog = , ,24
loglog wx = and40
loglog wy = .
12log40
log24
loglog
logloglogloglog =
++=
++=
twww
wzyx
wwxyz
Or 1211
401
241
=++t
.
Solving for t, we get t = 60. Thus 60log =wz . Method 3:
.601
401
241
121logloglogloglog =−−=−−== yxxyz
xyzxyz wwwww
∴ . 60log =wz
Method 4:
By (4.3), we have ⇒ 12log =wxyz 121log =xyzw (1)
By (4.1): (1) becomes 121
loglog
loglog
loglog
=++wz
wy
wx ⇒
121
log1
log1
log1
=++www zyx
601
401
241
121
log1
=−−=wz
⇒ 60log =wz .
Example 24: (1984 AIME) Determine the value of ab if and
5loglog 248 =+ ba
.7loglog 248 =+ ab
Solution: 512. Method 1: Adding the two given equations, we get:
12loglog 2248 =+ baab
By the Change-of-Base formula, we can change the two logarithms from base 8 and 4,
respectively into base 2: 12log3
log2
2 =+ abab .
50 AMC Lectures Chapter 33 Logarithms
So ,12 9log2 =ablog34
2 =ab and ab = 29 = 512.
Method 2:
22
2488 loglogloglog baba +++ = 12loglog 22
48 =+ baab
22
332
)(log)(log 23 abab += 34
223
2 )(logloglog ababab =+=
Therefore 12)(log 34
2 =ab ⇒ 1234
2)( =ab ⇒ ab = 29 = 512.
245
50 AMC Lectures Chapter 33 Logarithms
PROBLEMS
Problem 1: (1967 AMC) Given ,log)(log)(log)(log xr
cq
bp
a=== all logarithms to the
same base and x ≠ 1. If yxacb
=2
, then y is:
(A) rp
q+
2
(B) q
rp2+ (C) 2q – p – r (D) 2q – pr (E) q2 – pr
Problem 2: (1966 AMC) If ,loglog MN NM = M ≠ N, MN > 0, M ≠ 1, N ≠ 1, then MN
equals: (A) 21 (B) 1 (C) 2 (D) 10 (E) a number greater than 2 and less than 10.
Problem 3: Simplify .32log)2log2)(log3log3(log 4
29384 −++ Problem 4: Solve 325loglog2 25 =+ xx . Problem 5: (1971 AMC) If ))(log(loglog))(log(loglog 243432 yx =
,0))(log(loglog 324 == z then the sum x + y + z is equal to (A) 50 (B) 58 (C) 89 (D) 111 (E) 1296 Problem 6: Find m if 16loglog8log4log 4843 =⋅⋅ m .
(A) .29 (B) 9. (C) 18. (D) 27
Problem 7: (1996 ARML) For integers x an y with 1 < x, y ≤ 100, compute the number of ordered pairs (x, y) such that . 3loglog 2 =+ xy yx
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50 AMC Lectures Chapter 33 Logarithms
Problem 8: Find if 66log44 a=3log2 , b=11log3 .
Problem 9: Simplify 21
10110 10loglog
101loglog 2
−⋅⋅⋅ aa aa .
Problem 10: (1989 NEAML) Determine the numerical value of: ( )( ))(log)(log 45
2 xy yx.
Problem 11: Solve . 10loglog 82 =+ xx Problem 12: Solve 6logloglog
3133 =++ xxx .
Problem 13: Solve
⎪⎩
⎪⎨
⎧
=++=++=++
(3). 2logloglog(2) 2logloglog(1) 2logloglog
16164
993
442
yxzxzyzyx
Problem 14: Solve . 5)312(log2 =−+ xx Problem 15: a, b, and c (c is the hypotenuse) are the lengths of three sides of a right triangle. Show that aaaa bcbcbcbc )()()()( loglog2loglog −+−+ ⋅=+ a ≠ 1.
Problem 16: Find if 45log4 a=10log3 and b=25log6 .
Problem 17: Find if ab ba loglog −3
10loglog =+ ab ba with a > b > 1.
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50 AMC Lectures Chapter 33 Logarithms
Problem 18: For positive numbers x and y, if xy = 490 and Problem 19: Compare 2x, 3y, and 5z if . x, y, z are positive numbers. zyx 532 == Problem 20: Find the greatest value of log x + log y if 2x + 5y = 20. Problem 21: Find the smallest value of (log x)2 + (log y)2 if xy2 = 100 with 1 ≤ x ≤ 10. Problem 22: (1989 AIME) Find if 2
2 )(log x ).(loglog)(loglog 2882 xx = Problem 23: (2000 AIME I) The system of equations 4))(log(log)2000(log 101010 =− yxxy
1))(log(log)2(log 101010 =− zyyz
0))(log(log)(log 101010 =− xzzx
has two solutions ( ) and ( ). Find 111 ,, zyx 222 ,, zyx .21 yy +
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50 AMC Lectures Chapter 33 Logarithms
SOLUTIONS TO PROBLEMS Problem 1: Solution: (C). Method 1: The first three given logarithmic equalities are equivalent to the following exponential equalities a = xp, b = xq, c = xr. Hence
yrpqrp
q
xxxx
acb
=== −−+
222
, y = 2q – p – r.
Method 2: We may express the relationship involving y in logarithmic form:
.logloglog2log cabxy −−= Substituting the given first relations for the logarithms on the right hand side of the equation yields y log x = 2q log x − p log x − r log x. Since x ≠ 1 and log x ≠ 0, division by log x yields y = 2q – p – r. Problem 2: Solution: (B). The identity together with the given equation yields
1))(log(log =NM MN
.12 =)(log MN
∴ 1 or – 1. If log =MN 1log =MN
.1−=
, then M = N, which is ruled out, so we may
conclude that log MN
∴ , MN = 1. 1−= NM Problem 3: Solution: 0.
429384 32log)2log2)(log3log3(log −++
45
23333
2log)2log212)(log
2log31
2log21( −++= = 0
45
61
41
31
21
=−+++ .
Problem 4: Solution: 5 and 25. We may rewrite 325loglog2 25 =+ xx as
249
50 AMC Lectures Chapter 33 Logarithms
3log
1log225
25 =+x
x ⇒ ⇒
01log3log2 25225 =+− xx
0)1)(log1log2( 2525 =−− xx
So ,21log25 =x 5252
1
1 ==x ; ,1log25 =x 252 =x .
We can check to see that these two values are both solutions by plugging them into the given equation. Problem 5: Solution: (C). Since for any base b ≠ 0, 0log =Nb only if N = 1, the given equations yield
.1)(loglog)(loglog)(loglog 322443 === zyx
,3log4 =x ,4log2
Moreover, since , only if
M = b, we have
1log =Mb
=y ,2log3 =z or equivalently x = 43, y = 24, z = 32. Adding these results gives x + y + z = 43 + 24 + 32 = 64 + 16 + 9 = 89. Problem 6: Solution: 9.
28log
log4log8log
3log4log
=⋅⋅m ⇒ 9log3log2log ==m ⇒ m = 9.
Problem 7: Solution: 108.
3loglog 2 =+ xy yx ⇒ 3log2log =+ xy yx ⇒ 3log
2log =+y
yx
x ⇒
⇒ ⇒ .
y (log x
0)2 =yx )(log 2 +
1(log −yx
xlog32 =)(log −yx
02log3)2 =+− yy x
Thus or 01log =−yx 02log =−yx . The solutions to the first equation are the 99 ordered pairs from (2, 2) to (100, 100); the solutions to the second equation are the 9 ordered pairs (2, 4), (3, 9), (4, 16),…,(10, 100). Thus there are 99 + 9 = 108 ordered pairs of solutions.
Problem 8: Solution: 1+ a + ab
2 + ab.
We have 11log2log211log12log
11log2log11log3log2log
44log66log66log
33
33
32
3
333
3
344 +
++=
+++
==
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50 AMC Lectures Chapter 33 Logarithms
and ⇒ a=3log2 a13log2 =
Therefore ab
aba
ba
ba
+++
=+⋅
++=
21
12
11
66log44 .
Problem 9: Solution: – ½.
21010 loglog2log aaa ==
22
log110log
101log 2 aaa −=−=
aaa logloglog1)
101(
101 −=−=
−
aaa log2110log
2110log 2
1
−=−=−
.
Therefore
21
101
210 10loglog
101loglog
−⋅⋅⋅ aa aa
21)
log21)(log)(
log1(log 2
2 −=−−−=a
aa
a .
Problem 10: Solution (official solution): ( )( ))(log)(log 45
2 xy yx
= ( )(log)(log21 45 xy yx ⎟
⎠⎞
⎜⎝⎛ ) (Step 1)
= ( )( xy yx loglog)45(21
⋅⎟⎠⎞
⎜⎝⎛ ) (Step 2)
10loglog
loglog10 =⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
yx
xy .
Note: We believe that step 1 is incorrect. In the given expression ( )( ))(log)(log 45
2 xy yx, x
can be either positive or negative. In step 1, x was forced to be positive in ⎟⎠⎞
⎜⎝⎛ )(log
21 5yx .
However, it is okay set ( ) )(log5)(log 225 yy xx = because y must be positive since y is the
base in the expression ( ))(log 4xy .
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50 AMC Lectures Chapter 33 Logarithms
Problem 11: Solution: Method 1 (incorrect way): By Law 3: the Power Identity, loga xr
= r loga x, the given equation can be written as ⇒ 10 log x = 10 ⇒ log x = 1. 10log8log2 =+ xxTherefore, the solution is x = 10. We can plug this back into the given equation to check and see that it is a solution. However, we missed one solution because we misused the law 3, since we are not allowed to move “r” to the front if x > 0. Method 2 (correct way): We know that x2 and x8 are positive, so by Law 1: the Product Identity log a x + log a y
= loga xy , the given equation becomes log( ⇒ x10 = 1010 ⇒ x = ± 10. 10)82 =⋅ xxPlugging these values back into the given equation, we can check and see that both values of x are the solutions. Problem 12: Solution: 27.
We know that bmnb a
mam loglog = .
∴ ,loglog 233 xx = 1
3331 logloglog 1
−== − xxx .
The given equation becomes ⇒ 6logloglog 13
233 =++ −xxx 6log 2
3 =x
⇒ x2 = 36 ⇒ x = ± 27 . x = – 27 is extraneous, so the value for x is x = 27.
Problem 13: Solution: 32
=x , 827
=y , 3
32=z .
We know that x > 0, y > 0, and z > 0. Use the Change-of-Base formula, we can change the bases to 2, 3, and 4 in (1), (2), and (3), respectively:
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50 AMC Lectures Chapter 33 Logarithms
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
=++
=++
=++
16.loglog21log
21log
9loglog21log
21log
4,loglog21log
21log
4444
3333
2222
yxz
xzy
zyx
⇒
⎪⎪⎩
⎪⎪⎨
⎧
=
=
=
(6). 16
(5), 9
(4), 4
xyz
zxy
yzx
Multiplication of all three equations together gives (xyz)2 = 242. Since x > 0, y > 0, z > 0, xyz = 24 (7) Square both sides of (4): x2yz = 16 (8)
Dividing equation (8) by equation (7), we get 32
=x .
Similarly, we can obtain the values of y and z:827
=y and 3
32=z .
Plugging these values back into the given equation, we can check and see that the values
32
=x , 827
=y , 3
32=z are the solutions.
Problem 14: Solution: 5. The given equation can be written as or 5)312(log2log 22 =−+ xx 5)]312(2[log2 =−xx
Therefore . Factoring, we get (2x + 1)(2x – 32) = 0. 52 2)2(31)2( =− xx
Since 2x + 1 > 0, we have 2x – 32 = 0. The answer is x = 5. Problem 15: Solution: Because a, b, and c are the lengths of three sides of a right triangle and c is the hypotenuse, rearranging the terms in the Pythagorean Theorem, we get
Substituting (4) and (5) into (1) yields .)2(24345log4 ba
bab−
++=
Problem 17: Solution: −83
.
Method 1: ( )2 = ab ba loglog − abab baba loglog4)log(log 2 ⋅−+
9644)
310( 2 =−= .
We know that a > b > 1, so ab ba log1log << ⇒ 0loglog <− ab ba .
38loglog −=− ab ba .
Method 2: Since a > b > 1, . 1log >ab
We also know that 3
10loglog =+ ab ba ⇒ 313
log1log +=+
aa
bb .
∴ ,3log =ab ,31log =ba 3
8331loglog −=−=− ab ba .
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50 AMC Lectures Chapter 33 Logarithms
Problem 18: Solution: 8. xy = 490, x > 0, y > 0 (1)
255
1∴ 7log2loglog +=+ yx
1)7log(log)7log(log =−+− yx (2)
We also have 4
143)7log(log)7log(log −=−+− yx (3)
From (2) and (3), we know that 7loglog −x and 7loglog −y are the two roots of
04
1432 =−− tt .
0)2
13)(2
11( =−+ tt .
Let x > y, 2
137loglog =−= xt .
∴ 5.67log2
137loglog +=+=x (4)
Since 1071021
<< , then .17log5.0 << From (4) we have 5.7log7 << x , so the integer part is 8. Problem 19: Solution: 5z > 2x > 3y. Since , we have zyx 532 == .5log3log2log kzyx ===
We know that x, y, z > 0, ∴ k > 0.
We have ,2log
kx = ,3log
kx = 5log
kz = .
∴ 03log2log8log9log32 >
−=− kyx ⇒ 2x > 3y (1)
05log2log32log25log52 <
−=− kzx ⇒ 2x < 5z (2)
Therefore 5z > 2x > 3y.
50 AMC Lectures Chapter 33 Logarithms
Problem 20: Solution: 1. We know that x > 0, y > 0, so log x + log y = log (xy). log x + log y will be the greatest if xy is the greatest. From AM-GM inequality, we have
xyyxyx 1025225220 =⋅≥+= Simplifying this inequality gives us xy ≤ 10, and so the greatest value of xy is 10 (when x = 5 and y = 2). Thus, the greatest value of log (xy) is 1 and the greatest value of log x + log y is 1.
Problem 21: Solution: 54 .
We know that y > 0 and x
y 10= .
222
222 )log211()(log10log)(log)(log)(log xx
xxyx −+=⎟
⎠
⎞⎜⎝
⎛+=+
54
52log
451log)(log
45 2
2 +⎟⎠⎞
⎜⎝⎛ −=+−= xxx
Since 1 ≤ x ≤ 10, then 0 ≤ log x ≤ 1.
When 52log =x , or 5
2
10=x , (log x)2 + (log y)2 has the smallest value of 54 .
Problem 22: Solution: 27.
Since xxxx
28 log31
2log31
8log1log === ,
)(loglog31)(loglog 2228 xx = (1)
Let . xy 2log=
(1) becomes: yy22 log
31
3log = ⇒ yy
23
2 log)3
(log = ⇒ yy=3)
3(
Rearranging the terms, we get . Since 0)27( 2 =−yy 0≠y , . 27)(log 22
2 == xy
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50 AMC Lectures Chapter 33 Logarithms
257
z10
,2,2
,2,2
20
Problem 23: Solution: 25. Let u , , . x10 v log=log= y10 w log=
We rewrite the given system of equations as
or ⎪⎩
⎪⎨
⎧
=+−−=+−−
=+−−
.11log1
log1
10
10
uwwuwvvwvuuv
⎪⎩
⎪⎨
⎧
=−−=−−=−−
.1)1)(1(log)1)(1(log)1)(1(
10
10
uwwvvu
From the first two equations, we get u = w. From the third equation we have two cases: u = v = 2 or u = v = 0. Case I: . This gives us log10=v ,1001 =x 201 =y , 1001 =z .
Case II: . This gives us 510log=v ,12 =x 52 =y , 12 =z . Therefore y1 + y2 = 25.