I suppose the equation should read
[tex]13^{x+7}=12^{-9x}[/tex]
Take the logarithm of both sides; the base of the logarithm doesn't really matter, so I'll make the "natural" choice:
[tex]\ln 13^{x+7}=\ln12^{-9x}[/tex]
Use the exponent property of logarithms:
[tex](x+7)\ln13=-9x\ln12[/tex]
Solve for [tex]x[/tex]:
[tex]x\ln13+7\ln13=-9x\ln12[/tex]
[tex]x(\ln13+9\ln12)=-7\ln13[/tex]
[tex]x=-\dfrac{7\ln13}{\ln13+9\ln12}[/tex]
Then divide through the numerator and denominator by [tex]\ln13[/tex]:
[tex]x=-\dfrac7{1+9\frac{\ln12}{\ln13}}[/tex]
Use the change of base formula to rewrite
[tex]\dfrac{\ln12}{\ln13}=\log_{13}12[/tex]
So we end up with one of many ways of expressing the solution:
[tex]\boxed{x=-\dfrac7{1+9\log_{13}12}}[/tex]
