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

Dec 12, 2016

$n = \frac{2 \ln \left(3\right)}{2 \ln \left(2\right) + \ln \left(3\right)}$

Explanation:

There are a variety of methods, but I'd like to start by squaring both sides to undo the square root.

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

The left hand side will use ${\left({a}^{b}\right)}^{c} = {a}^{b c}$:

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

Typically, to solve an exponential such as this, we want to take the logarithm with a base of whatever the base of the exponential is. However, since we have two bases here, we can use an arbitrary logarithm. A common logarithm to use, although it doesn't really matter, is the natural logarithm $\ln \left(x\right)$ which has a base of $e$.

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

Rewriting these using $\ln \left({a}^{b}\right) = b \ln \left(a\right)$ gives:

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

Expanding the right hand side:

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

Grouping the terms with $n$ and factoring:

$2 n \ln \left(2\right) + n \ln \left(3\right) = 2 \ln \left(3\right)$

$n \left(2 \ln \left(2\right) + \ln \left(3\right)\right) = 2 \ln \left(3\right)$

Solving:

$n = \frac{2 \ln \left(3\right)}{2 \ln \left(2\right) + \ln \left(3\right)}$