Log question help

I can't seem to find x although Im pretty sure that I've done it right. Can you look at the question and tell me if there is an answer or not please

Thank you :smile:

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Posted from TSR Mobile

Reply 1

Pariah

With a difficult log don't push things or you'll get piles

Reply 2

ChildishHambino

Original post by puddinboy

I can't seem to find x although Im pretty sure that I've done it right. Can you look at the question and tell me if there is an answer or not please

Thank you :smile:

Posted from TSR Mobile

Move them to seperate sides of the the equals sign, inverse log either one, recognize that 4^2 = 16, use laws of indices with log rules and you should be left with a quadratic in x.

Reply 3

3.1415927

What I did was:

(log16(x))/(log16(4))= log16(/x-12-/)

=log16(x) = 2log16(/x-12/)

therefore x=(/x-12/)^2

thus, x=x^2 -24x + 144

then x^2-25+144
Finally (x-16)(x-9)
so yeah x=9 or x=16 .. I think