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