# Logarithms Quiz Set 017

### Question 1

What is x if logx$(3/5)$ = $1/2$?

A

${9/25}$.

B

${3/5}$.

C

$√{3/5}$.

D

$√{5/3}$.

Soln.
Ans: a

logx$(3/5)$ = $1/2$, by definition, gives $x^{1/2}$ = $3/5$. So x = $(3/5)^2$ = ${9/25}$.

### Question 2

What is log$4$ + log$1/4$?

A

0.

B

$\text"log"_3(4)$.

C

$\text"log"_4(3)$.

D

1/2.

Soln.
Ans: a

log(m) + log(1/m) = log (m × $1/m$) = log 1 = 0.

### Question 3

What is the value of log2(128)?

A

7.

B

$√7$.

C

0.

D

$√128$.

Soln.
Ans: a

Let log2(128) = P. By definition, we have 2P = 128 = 27, which gives 7 as the answer.

### Question 4

What is P if $\text"log"_5(4)$ + $\text"log"_5(4P + 1)$ = 1 + $\text"log"_5(P + 4)$?

A

$1{5/11}$.

B

$2{7/10}$.

C

${5/13}$.

D

$3{10/13}$.

Soln.
Ans: a

We have $\text"log"_5(4)$ + $\text"log"_5(4P + 1)$ = $\text"log"_5(5)$ + $\text"log"_5(P + 4)$. It is same as $\text"log"_5(4 × (4P + 1))$ = $\text"log"_5(5 × (P + 4))$. Equating the logs, $4 × (4P + 1) = 5 × (P + 4)$, solving for P we get P = $1{5/11}$.

### Question 5

What is the value of log2(128)?

A

7.

B

$√7$.

C

0.

D

$√128$.

Soln.
Ans: a

Let log2(128) = P. By definition, we have 2P = 128 = 27, which gives 7 as the answer.