<PRE><B>Question 536128</B>
Remember that log and exponents are inverse such that:&lt;P&gt;
If b^y=x then log_b(x) = y.&lt;P&gt;
5^(x-7)=9^(-9x) can be rewritten as:  log_5(9^(-9x))=x-7&lt;P&gt;
Remember log (x^y) = y*log(x).&lt;P&gt;

Rewrite as:  (-9x)*log_5(9)=x-7&lt;P&gt;
Convert log_x to log_y with log_y(z) = log_x(z)/log_x(y)&lt;P&gt;
We can convert from base-5 to base-10 in this case.&lt;P&gt;
That gives:  (-9x)*(log_10(9)/log_10(5)) = x-7.&lt;P&gt;
Now evaluate the logs and complete the algebra.&lt;P&gt;

-9x*(1.36521239)=x-7  ---&gt; -12.2869115x=x-7  (subtract x from both sides)&lt;P&gt;

-13.2869115x=-7  (divide both sides by -13.2869115)  x=0.526834246

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