January 13, 2015
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
4 - Logarithms
1. Given that log N = log 31 + log 13 + 1 − log 2, compute N .
2. For positive integers x and y, xy = y x = 16. Compute all possible values of
logx y + logy x
.
x−y
3. If log 12 = A and log 12500 = B, find log 15 in terms of A and B in simplest form.
January 2015
4 - Logarithms
1. Given that log N = log 31 + log 13 + 1 − log 2, compute N .
SOLUTION: 2015 We have
log N = log 31 + log 13 + 1 − log 2 = log 31 + log 13 + log 10 − log 2 = log
31 · 13 · 10
, or
2
log 2015.
2. For positive integers x and y, xy = y x = 16. Compute all possible values of
logx y + logy x
.
x−y
SOLUTION: ±1.25or± 54 Since 24 = 16 and 42 = 16, we have (x, y) = (2, 4) or (4, 2).
logx y + logy x
2 + 21
log2 4 + log4 2
=
, which equals
= ±1.25.
Therefore,
x−y
±2
±2
3. If log 12 = A and log 12500 = B, find log 15 in terms of A and B in simplest form.
SOLUTION: A + B − 4 Note that A = log 3 + log 4 and B = log 5 + log 2500. Now,
log 15 = log 3 + log 5 = A + B − log 4 − log 2500, or A + B − 4.
January 8, 2014
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
4 - Logarithms
1. Compute log4 2 + log3 3 + log2 4.
2. Compute log2 36 · log6 8.
x+2
3. Suppose that y1 = 2(4x ) and that y2 = 8 4 . For some x = a log2 b + c, where a, b, and c
are integers and b is as small as possible, y2 = 81y1 . Compute the ordered triple (a, b, c).
January 2014
4 - Logarithms
1. Compute log4 2 + log3 3 + log2 4.
SOLUTION: 3.5 or 72 We compute the values of a, b, and c such that 4a = 2, 3b = 3,
and 2c = 4. Our answer is
1
2
+ 1 + 2 = 27 .
2. Compute log2 36 · log6 8.
SOLUTION: 6 This simplifies using the change of base rule: log2 36 · log6 8 =
36 log 8
· log 2 , which is log6 36 · log2 8 = 2 · 3 = 6.
which is equivalent to log
log 6
log 36
log 2
·
x+2
log 8
,
log 6
3. Suppose that y1 = 2(4x ) and that y2 = 8 4 . For some x = a log2 b + c, where a, b, and c
are integers and b is as small as possible, y2 = 81y1 . Compute the ordered triple (a, b, c).
SOLUTION: (4, 3, −3) Note that y2 = 81y1 = 81(2(4x )) = 81(22x+1 ). Also,
x+2
3x+6
y2 = 8 4 = 2 22 = 23x+4 . Thus, 81(22x+1 ) = 23x+4 → 81 = 2x+3 . This means that
x + 3 = log2 81 = 2 log2 9 = 4 log2 3. If b is as small as possible, x = 4 log2 3 − 3, so our
ordered pair is (4, 3, −3).
January 9, 2013
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
4 - Logarithms
1. Compute (log2 16)(log9 27) .
2. Let P denote a point represented by the coordinates of an x-intercept of
y = f (x) = log2 (8(2x − 1)2 ) − 5. Let Q denote a point represented by the coordinates of a
y-intercept of this function. Compute all possible distances P Q.
3. Given A = log10 2 + log10 32 + log10 34 + · · · + log10
simplified single fraction in terms of k.
2000
1999
and log1024 10 = k. Express A as a
January 2013
4 - Logarithms
1. Compute (log2 16)(log9 27) .
√ 3
3
SOLUTION: 8 This is 4 2 = 4 = 8.
2. Let P denote a point represented by the coordinates of an x-intercept of
y = f (x) = log2 (8(2x − 1)2 ) − 5. Let Q denote a point represented by the coordinates of a
y-intercept of this function. Compute all possible distances P Q.
√
SOLUTION: 52 and 21 17 x = 0 → y = 3 − 5 = −2.
y = 0 → log2 (8(2x − 1)2 ) = 5 → 8(2x − 1)2 = 25 = 32, so 2x − 1 = ±2 ⇒ x = 32 or x = − 21 .
The distance between (0, −2) and (3/2, 0) is 25 . The distance between (0, −2) and (−1/2, 0)
√
is 12 17.
3. Given A = log10 2 + log10 32 + log10 34 + · · · + log10
simplified single fraction in terms of k.
SOLUTION:
A = log10 (2 · 23 ·
2000
1999
and log1024 10 = k. Express A as a
30k+1
Using the so-called Product Property of
10k
· · · · 2000
) = log10 (2000) = 3 + log10 2. We note
1999
1
log1024 (10) = k ⇒ k =
⇒ log10 2 =
log(210 )
1
.
10k
Thus, A = 3 +
Logs, we have
that
1
10k
=
30k+1
.
10k
January 11, 2012
Any calculator permitted on N.Y.S. Regents examinations may be used.
The word “compute” calls for an exact answer in simplest form.
4 - Logarithms
1. If log x = 2 log 4 − 3 log 2, compute x.
2. On a number line, the coordinates of A and C are log 50 and log 100, respectively. If the
coordinate of B, the midpoint of segment AC, is log x, compute x.
3. The solution to 52 log5 x − 12 4log2
Compute the solution in this form.
√ x
√
− 9log3 4 = 0 is a number of the form a + b c.
January 2012
4 - Logarithms
1. If log x = 2 log 4 − 3 log 2, compute x.
SOLUTION: 2 By the Power Property of Logs, we have log x = log(42 ) − log(23 ). Using the
Quotient Property of Logs, we have log x = log 16
8 , so x = 2.
2. On a number line, the coordinates of A and C are log 50 and log 100, respectively. If the
coordinate of B, the midpoint of segment AC, is log x, compute x.
√
√
100
= log 5000
= log 5000. This simplifies to
SOLUTION: 50 2 We compute log 50+log
2
2
√
√
log 50 2, so x = 50 2.
√ √
3. The solution to 52 log5 x − 12 4log2 x − 9log3 4 = 0 is a number of the form a + b c. Compute
the solution in this form.
√
2
2
SOLUTION: 6 + 2 13 The given equation is equivalent to 5log5 x − 12 4log4 x − 3log3 4 = 0.
√
√
12± 144−4(1)(−16)
12± 208
This is equivalent to x2 − 12x − 16 = 0, which has roots
=
. One of
2
√ 2
these roots is negative, so we reject it. The positive root simplifies to 6 + 2 13.
JANUARY TOPICS – February 2, 2011
CALCULATORS MAY BE USED ON ALL TOPICS
#4 LOGARITHMS
ANSWERS:
1.
−
1
2
2.
5a + 5
2
3.
x < −4 or 1 < x <
3
2
or x >


6
6
1. Compute: log 9 1 +
 + log 9 1 −

3 
3 


2. Given: log10
2
=a
5
Express log10 32 in terms of a, without making reference to any additional logarithms.
3. Determine the domain of the real-valued function f:
 x 2 + 3x − 4 
f ( x ) = log 3 

 ( 2 x − 3) 2 


3
2
JANUARY 14, 2010
CALCULATORS MAY BE USED ON ALL TOPICS
#4 LOGARITHMS
ANSWERS:
1.
4
2.
2
3.
⅔, 2
1. Evaluate: log 4 1152 − log 4 4 12
2. If A and B are the roots of 3 x 2 − 22 x + 27 = 0, then what is the
exact value of log 3 A + log 3 B ?
3. Compute all possible real values of x for which 31+5log3 x − 40.5+ 4log4 x + 23+ 2log2 x = 12 x3 .
January 2010
JANUARY 8, 2009
CALCULATORS MAY BE USED ON ALL TOPICS
#4
LOGARITHMS
ANSWERS:
1.
14
3
2.
8
3.
½, 1
1. Compute the value of: log8 4 − log 13 81
2. Find all real solutions for: log 2 x + log 2 ( x − 6 ) = 4
3. Compute all possible values of x for which 33log3 x +1 − 22log 2 x = 2 x 4 .