# How do you solve 2^(x+1)=3^x?

Feb 17, 2015

I would write it as:
${2}^{x} \cdot {2}^{1} = {3}^{x}$
$\frac{{2}^{x}}{{3}^{x}} = \frac{1}{2}$
${\left(\frac{2}{3}\right)}^{x} = \frac{1}{2}$
take the log of both sides:
$\ln {\left(\frac{2}{3}\right)}^{x} = \ln \left(\frac{1}{2}\right)$
$x \ln \left(\frac{2}{3}\right) = \ln \left(\frac{1}{2}\right)$
$x = 1.71$

Feb 19, 2015

The answer is: $x = \ln \frac{2}{\ln 3 - \ln 2}$.

${2}^{x + 1} = {3}^{x} \Rightarrow \ln {2}^{x + 1} = \ln {3}^{x} \Rightarrow \left(x + 1\right) \ln 2 = x \ln 3 \Rightarrow$

$x \ln 2 + \ln 2 = x \ln 3 \Rightarrow x \left(\ln 2 - \ln 3\right) = - \ln 2 \Rightarrow x = - \ln \frac{2}{\ln 2 - \ln 3} = \ln \frac{2}{\ln 3 - \ln 2}$.