Question 978888
 *[illustration V14.JPG].
1. First do a conversion of the equation.
Bring the linear term into the logarithm. 
So then it looks like,
{{{log((3^x-6),sqrt(3)^(6-2x))<=1}}}
Convert to log base 10,
{{{log(10,(sqrt(3)^((6-2x))))/log(10,(3^x-6))<=1}}}
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From the graph you can see there are three regions of interest. 
They are limited by the endpoints where,
{{{3^x-6=0}}}
{{{3^x=6}}}
{{{x=ln(6)/ln(3)}}}
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{{{3^x-6=1}}}
{{{3^x=7}}}
{{{x=ln(7)/ln(3)}}}
So the function is less than 1 when,
{{{highlight(ln(6)/ln(3)<x<ln(7)/ln(3))}}}
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The other endpoint is determined when, 
{{{3^x-6=3^(x-3)}}}
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{{{log(10,(3^((3-x))))/log(10,(3^x-6))<=1}}}
{{{log(10,(3^((3-x))))<=log(10,(3^x-6))}}}
{{{3^((3-x))<=3^x-6}}}
{{{27*3^(-x)<=3^x-6}}}
{{{27<=3^(2x)-6^(x)}}}
Which converts to a quadratic inequality,
{{{y^2-6y-27>=0}}}
{{{(y-9)(y+3)>0}}}
with solutions {{{y>=9}}} and {{{y<=-3}}}
We can only use the positive solution.
{{{y>=9}}}
{{{3^x>=9}}}
{{{highlight(x>=2)}}}

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The second problem is solved following the same procedure.