How do you solve 2^x=sqrt(3^(x-2))?

Nov 8, 2016

Square both sides.

${\left({2}^{x}\right)}^{2} = {\left(\sqrt{{3}^{x - 2}}\right)}^{2}$

${2}^{2 x} = {3}^{x - 2}$

Take the natural logarithm of both sides.

$\ln \left({2}^{2 x}\right) = \ln \left({3}^{x - 2}\right)$

Use the power rule of logarithms, $\ln \left({a}^{n}\right) = n \ln a$

$\left(2 x\right) \ln 2 = \left(x - 2\right) \ln 3$

$2 x \ln 2 = x \ln 3 - 2 \ln 3$

$2 x \ln 2 - x \ln 3 = - 2 \ln 3$

$x \left(2 \ln 2 - \ln 3\right) = - 2 \ln 3$

You can simplify the expression in parentheses using the rules $n \ln a = \ln {a}^{n}$ and $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$.

$x \left(\ln 4 - \ln 3\right) = - 2 \ln 3$

$x \left(\ln \frac{4}{3}\right) = \ln \left(\frac{1}{9}\right)$

$x = \ln \frac{\frac{1}{9}}{\ln \left(\frac{4}{3}\right)}$

$x \cong - 7.64$

Hopefully this helps!