Question

How do you solve for x in `2^x = 5^(x - 2)`?

Answer 1

I found: `x=(2ln(5))/(ln(5)-ln(2))`

Explanation:

I would apply the natural logarithm, `ln`, to both sides:
`ln(2^x)=ln(5^(x-2))`
apply one of the rules of logs:
`xln(2)=(x-2)ln(5)`
`xln(2)=xln(5)-2ln(5)` rearranging:
`x[ln(5)-ln(2)]=2ln(5)`
and finally:
`x=(2ln(5))/(ln(5)-ln(2))`

Using a pocket calculator you can get `x=3.513`