Question

express log2^3 -log64+1/2log16 as a single logarithm

Answer 1

log 8 - log 8^2 + log √16
log 8 - 2 log 8 + log 4
- log8 + log 4

log 4/8

log 1/2
- log2

Answer 2

Answer:

$log2^3-log64+\frac{1}{2}log16=log\frac{1}{2}=-log2$

Step-by-step explanation:

Given :  $log2^3-log64+\frac{1}{2}log16$

To Express: In single logarithm

Solution : $log2^3-log64+\frac{1}{2}log16$

$=log8-log8^2+log\sqrt{16}$  

$=log8-2log8+log4$  

$=-log8+log4\rightarrow log4-log8$

Logarithm property  $logA-logB=log\frac{A}{B}$

$log4-log8=log\frac{4}{8}=log\frac{1}{2}=log2^{-1}=-log2$

Therefore,  $log2^3-log64+\frac{1}{2}log16=log\frac{1}{2}=-log2$