Question

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

Answer 1

`n=(2ln(3))/(2ln(2)+ln(3))`

Explanation:

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

`(2^n)^2=(sqrt(3^(n-2)))^2`

The left hand side will use `(a^b)^c=a^(bc)`:

`2^(2n)=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(x)` which has a base of `e`.

`ln(2^(2n))=ln(3^(2-n))`

Rewriting these using `ln(a^b)=bln(a)` gives:

`2nln(2)=(2-n)ln(3)`

Expanding the right hand side:

`2nln(2)=2ln(3)-nln(3)`

Grouping the terms with `n` and factoring:

`2nln(2)+nln(3)=2ln(3)`

`n(2ln(2)+ln(3))=2ln(3)`

Solving:

`n=(2ln(3))/(2ln(2)+ln(3))`