SOLUTION: how do you write 5^x-7=9^-9x as a 10 logrithm finding the value of x. Thanks for your help

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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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You can put this solution on YOUR website!
how do you write 5^x-7=9^-9x as a 10 logrithm finding the value of x.
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5^(x-7)=9^-9x
take log of both sides using exponent rule
(x-7) log5=-9xlog9
(x-7)/9x=-log9/log5
using calculator
(x-7)/9x=-1.3652
x-7=9x*-1.362=-12.2869x
x+12.2869x=7
13.2869x=7
x=7/13.2869≈.5268