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9^(-x+9)= 11^(-5x) solve for x.
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\[\ln 9^{(-x+9)}=\ln 11^{(-5x)}\] \[(-x+9) \ln9=-5x \ln11\] \[-x*\ln9+9*\ln9=-5x*\ln11\] get x's together \[-x*\ln9+5x*\ln11=-9 \ln9\] \[x(-\ln9+5\ln11)=-9\ln9\] get x by itself by dividing by -ln9+5ln11
\[x=\frac{-9\ln9}{5\ln11-\ln9}\]
would it be the same answer if we put \[9\log (9)\div \log (9) - 5\log (9)\] ??
no you could write \[x=\frac{9\ln9}{\ln9-5\ln11}\]
is that what you meant?
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