How do you solve \log _ { 8} 2+ \log _ { 8} 2x ^ { 2} = \log _ { 8} 64log82+log82x2=log864?

1 Answer
Jan 16, 2018

x=+-4x=±4

Explanation:

log_8 2+log_8 2x^2=log_8 64log82+log82x2=log864

Rearrange:

->log_8 2x^2=log_8 64 - log_8 2log82x2=log864log82

Using the rules of logs:

log_8 2x^2=log_8 (64/2)log82x2=log8(642)

log_8 2x^2=log_8 (32)log82x2=log8(32)

So:

8^(log_8 2x^2)=8^(log_8 (32))8log82x2=8log8(32)

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

->2x^2=32

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