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

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Answer 1 · 2018-01-16T17:09:16

#x=+-4#

#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#