Question

How do you solve `4^(p-1) ≤ 3^p`?

Answer 1

The answer is : `p<=log_{4/3}(3)-1`

Explanation:

We start from the inequality :

`4^{p-1}<=3^p`

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

`4^{p-1}<=3*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:

`(4/3)^{p-1}<=3`

Now we can write the right side as the exponent using rule which says that: `a=b^{log_a b}`

We get then:

`(4/3)^{p-1}<=(4/3)^(log_{4/3}(3))`

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

`p-1<=log_{4/3}(3)` (*)

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

`p<=log_{4/3}(3)+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 `<=` to `>=` when changing from exponential inequality to inequality of exponents.