Question

How do you solve for x in `18^(x-2) = 13^(-2x)`?

Answer 1

`x = (2ln(18))/(ln(18) +2ln(13))`

Explanation:

Since 13 is a prime and 18 is a compound number without 13 as a prime factor you'll never be able to equate bases, therefore, we'll need to apply logs. I'm using the natural base due to preference you can use whatever base you want - that being said, I'd recommend using natural or decimal.

`ln(18^(x-2)) = ln(13^(-2x))`
`(x-2)ln(18) = (-2x)ln(13)`
`xln(18) -2ln(18) = -2xln(13)`

Isolating x

`xln(18) +2xln(13) = 2ln(18)`
`x*(ln(18) + 2ln(13)) = 2ln(18)`
`x = (2ln(18))/(ln(18) +2ln(13))`

You could break down some of these logs further but since the problem gave 18 and 13 as numbers, I'd leave the answer in terms of their logs.