# How do you solve 4^(p-1) ≤ 3^p?

Apr 3, 2016

The answer is : $p \le {\log}_{\frac{4}{3}} \left(3\right) - 1$

#### Explanation:

We start from the inequality :

## ${4}^{p - 1} \le {3}^{p}$

If we rewrite the right side as ${3}^{p - 1} \cdot 3$ we will get the same exponents on both sides:

## ${4}^{p - 1} \le 3 \cdot {3}^{p - 1}$

Now if we divide both sides of the inequality by: ${3}^{p - 1}$ we will get the exponents only on the right side:

## ${\left(\frac{4}{3}\right)}^{p - 1} \le 3$

Now we can write the right side as the exponent using rule which says that: $a = {b}^{{\log}_{a} b}$

We get then:

## ${\left(\frac{4}{3}\right)}^{p - 1} \le {\left(\frac{4}{3}\right)}^{{\log}_{\frac{4}{3}} \left(3\right)}$

Now, when we have powers with the same bases we can write our inequality as inequality of the exponents:

## $p - 1 \le {\log}_{\frac{4}{3}} \left(3\right)$ (*)

Finally we can move $1$ to the right side to get the solution

## $p \le {\log}_{\frac{4}{3}} \left(3\right) + 1$ 

Note:

In the expression marked with (*) if the base was lower than 1 we would have to change the sign of the inequality fron $\le$ to $\ge$ when changing from exponential inequality to inequality of exponents.