How can logarithms be used to solve exponential equations?

[anonymous12]

Homework Statement

Solve $2^{k-2}=3^{k+1}$

Homework Equations

The Attempt at a Solution

$$2^{k-2}=3^{k+1}$$
$$\frac{2^{k}}{2^{2}}=(3^k)(3^1)$$
$$\frac{2^{k}}{3^{k}} = 4 \cdot 3$$
$$\frac{2^{k}}{3^{k}} = 12$$

What do I do next to solve for K?

 

[Mentallic]

Use the fact that $$\frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n$$ and also, what do you know about logarithms?

 

[HallsofIvy]

You could have used logarithms right from the start- $log(2^{k-2})= log(3^{k+1})$.

In general, to solve an equation of the form f(x)= constant or f(p(x))= f(q(x)) you will need to use the inverse function to f. And the inverse of the exponential is the logarithm.