Question 101035
{{{8^(3x+4)=32^(1-x)}}}

This should be in LOGARITHMS!!!


Read my lesson---An overview to the laws of logarithms 

then, go back to this page


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{{{log(8^(3x+4))=log(32^(1-x))}}}
{{{3x+4(log(8))=1-x(log(32))}}}

In tead of a logarithm witha base of ten, it would be easier if we choose a base of 2 since {{{2^3=8}}} and {{{2^5=32}}}

{{{3x+4(3)=1-x(5)}}}
{{{3(3x+4)=5(-x+1)}}}
{{{9x+12=-5x+5}}}
{{{9x+5x+12=-5x+5x+5}}}
{{{14x+12=0+5}}}
{{{14x+12=5}}}
{{{14x+12-12=5-12}}}
{{{14x+0=-7}}}
{{{14x=-7}}}
{{{14x/14=-7/14}}}
{{{x=-1/2}}}


Power up,
HyperBrain!