Question

How do you solve `5^(x-1) = 2^x`?

Answer 1

I found: `x=(ln5)/(ln5-ln2)`

Explanation:

I would take the natural log (`ln`) on both sides:
`ln5^(x-1)=ln2^x`
then use the fact that:
`lnx^a=alnx`
to get:
`(x-1)ln5=xln2`
`xln5-ln5-xln2=0`
`x(ln5-ln2)=ln5`
so that:
`x=(ln5)/(ln5-ln2)`
This is equal to `=1.7564` (if you can use the calculator).