How do you solve #log_8(8/x^2)=3*(log_8x)^2#?

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Answer 1 · 2017-07-14T14:47:10

#x=2# or #x=1/8#

#log_8(8/x^2)=3*(log_8x)^2# can be expanded as

#log_8 8-2log_8x=3*(log_8x)^2#

or #1-2log_8x=3*(log_8x)^2#

Now assuming #u=log_8x# this becomes

#3u^2+2u-1=0#

or #(3u-1)(u+1)=0#

i.e. #u=1/3# or #u=-1#

If #u=1/3#, we have #log_8x=1/3# i.e. #8^(1/3)=x# or #x=2#

and if #u=-1#, we have #log_8x=-1# i.e. #8^(-1)=x# or #x=1/8#