Question 538256
{{{x=b^y}}} is the same as {{{y = log(b,x)}}}<P>
That's the definition of logarithms.
The problem gives {{{1/6=4^(7+3x)}}} which is the same as {{{7+3x = log(4,1/16)}}}<P>
Look at the right side. {{{log(4,1/16)}}}<P>
Back to the definition of logarithms.  Turn it into an equation because now we want to find its value.<P>
{{{y=log(4,1/16)}}} means {{{1/16 = 4^y}}}<P>
{{{1/(4^2)=1/16}}}<P>
Remember that {{{x^(-y) = 1/x^y}}}.<P>
So the answer is y=-2 because {{{4^-2 = 1/16}}}<P>
{{{-2=log(4,1/16)}}} so substitute -2 into the right side of the equation.<P>
{{{7+3x = -2}}}<P>
Subtract 7 from both sides, then divide both sides by 3.<P>
{{{x=-3}}}<P>
Check the original equation.<P>
{{{4^(7+3(-3))= 4^(7-9) = 4^-2 = 1/16}}}
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