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PRACTICE PROBLEMS
PARK, BAE JUN
Logarithm
Math114
Section302 & 308
(1) Write the expression as a single logarithm with a coefficient of 1.
1
log5 6 + log5 + log5 10
3
log5 (6 · 13 · 10) = log5 20
∴ log5 20
(2) Simplify logb (3b4 ).(Circle only one answer)
A. 1 + 4 logb 3
B. 4 + 4 logb 3
C. 4 + logb 3
D. 12
logb (3b4 ) = logb 3 + logb b4 = logb 3 + 4 logb b = logb 3 + 4
∴C
1
E. none of these
2
PARK, BAE JUN
(3) Given A = log3 x and log3 y, express the followings as a single logarithm with a
coefficient of 1.
(a)2A + 3B
2A + 3B = 2 log3 x + 3 log3 y = log3 x2 + log3 y 3 = log3 (x2 y 3 )
∴ log3 (x2 y 3 )
(b)3A − 2B
3A − 2B = 3 log3 x − 2 log3 y = log3 x3 − log3 y 2 = log3
∴ log3
(c)
A
B
x3
y2
A
B
=
log3 x
log3 y
∴ logy x
= logy x
x3
y2
SECTION 302 & 308
3
(4) Which of the following is equal to log 7? Justify your answer.
a.
log2 10
log2 7
b.
log2 7
log2 10
c.
log2 7
log 2
d.
log 2
log2 7
e.
log7 10
log2 10
f.
log2 10
log7 10
a.log7 10
b.log10 7 = log 7
log 7
log 7
log 7
log 2
c. 2 =
=
log 2
log 2
(log 2)2
log 2
(log 2)2
d. log 7 =
log 7
log 2
1
log 7
e. 1
log 2
1
log 2
f. 1
log 7
=
log 2
= log7 2
log 7
=
log 7
= log2 7
log 2
∴b
(5) Which of the following is equal to log 2? Justify your answer.
a.
a.
log 3
log2 3
log 3
log 3
log 2
1
log 2
b. 1
log 3
1
log 3
c. 1
log 2
b.
log2 10
log3 10
c.
= log 2
=
log 3
= log2 3
log 2
=
log 2
= log3 2
log 3
d.log3 10
e.log10 3 = log 3
f.
log 3
log 2
log 3
∴a
=
1
= log2 10
log 2
log3 10
log2 10
d.
log2 10
log2 3
e.
log2 3
log2 10
f.
log2 3
log 3
4
PARK, BAE JUN
(6) Find a number t such that log2 t = −5.
t = 2−5 =
1
32
(7) Given log4 x = 3.2 and log4 y = 1.3, evaluate log4 x3 y 4 .
log4 x3 y 4 = log4 x3 + log4 y 4 = 3 log4 x + 4 log4 y = 3 · 3.2 + 4 · 1.3 = 14.8
(8) Find all numbers x that satisfy the equation log5 (x + 6) + log5 (x + 2) = 1.
log5 (x + 6)(x + 2) = 1 and x + 6 > 0, x + 2 > 0
⇒ (x + 6)(x + 2) = 51 = 5 and x > −2
⇒ x2 + 8x + 12 = 5 ⇒ x2 + 8x + 7 = 0 ⇒ (x + 1)(x + 7) = 0 ⇒ x = −1, −7
Since x > −2, x = −1.
∴ x = −1
SECTION 302 & 308
(9) Find all numbers x that satisfy the equation log9 (x + 1) =
5
1
+ log9 x.
2
log9 (x + 1) = log9 3 + log9 x = log9 (3x)
⇒ (x + 1) = 3x and x + 1 > 0, 3x > 0
1
2
1
∴x=
2
⇒x=
(10) Find all numbers x that satisfy the equation log2 (x + 16) − 1 = log2 (x − 2).
log2 (x + 16) − log2 2 = log2 (x − 2)
⇒ log2
⇒
x+16
2
= log2 (x − 2)
x + 16
= x − 2 and x > −16, x > 2
2
⇒ x + 16 = 2x − 4 ⇒ x = 20
∴ x = 20