How do you solve \log _ { 8} 2+ \log _ { 8} 2x ^ { 2} = \log _ { 8} 64?

1 Answer
Jan 16, 2018

x=+-4

Explanation:

log_8 2+log_8 2x^2=log_8 64

Rearrange:

->log_8 2x^2=log_8 64 - log_8 2

Using the rules of logs:

log_8 2x^2=log_8 (64/2)

log_8 2x^2=log_8 (32)

So:

8^(log_8 2x^2)=8^(log_8 (32))

cancel8^(cancel(log_8) 2x^2)=cancel8^(cancel(log_8) (32))

->2x^2=32

x^2=16-> x=+-4