How do you solve 92^x = 27^(x - 1)?

Nov 14, 2016

Explanation:

Take a logarithm of any base of both sides (I am going use the natural logarithm):

$\ln \left({92}^{x}\right) = \ln \left({27}^{x - 1}\right)$

Use the property of all logarathims ${\log}_{b} \left({a}^{c}\right) = \left(c\right) {\log}_{b} \left(a\right)$

$\left(x\right) \ln \left(92\right) = \left(x - 1\right) \ln \left(27\right)$

Use the distributive property:

$\left(x\right) \ln \left(92\right) = \left(x\right) \ln \left(27\right) - \ln \left(27\right)$

Solve for x:

$\left(x\right) \ln \left(92\right) - \left(x\right) \ln \left(27\right) = - \ln \left(27\right)$

$x = \ln \frac{27}{\ln \left(27\right) - \ln \left(92\right)}$

$x \approx - 2.68839$