Question

Evaluate 5log2(32)
A. 10
B. 25 C. 64/5 D. 64 E. 160

Answer 1

B

Explanation:

using the laws of logarithms

• log
`x^(n)` ⇔ nlogx

•
`log_(b)` x = n ⇒ x =
`b^(n)`

given

5
`log_(2)` (32)

=
`log_(2)`
`(32)^(5)` ← let this equal n , then solve for n, the value of the logarithm

`log_(2)`
`(32)^(5)` = n

`(32)^(5)` =
`2^(n)`

`(2^5)^(5)` =
`2^(n)`

`2^(5(5))` =
`2^(n)`

`2^(25)` =
`2^(n)`

since bases are same , both 2 then equate exponents

n = 25