Question

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

Answer 1

I found `x=1.7564` but....

Explanation:

I would try by....cheating a bit! I imagine that you can use a pocket calculator to evaluate the natural logarithm, `ln`. So, I take the natural log of both sides:
`ln(5)^(x-1)=ln(2)^x`
I use the rule of logs that tells us that: `loga^x=xloga` to get:
`(x-1)ln(5)=xln2`
`xln5-ln5=xln2` rearranging:
`x(ln5-ln2)=ln5`
`xln(5/2)=ln(5)`
and `x=ln5/(ln(5/2))=` here comes the "cheating" bit:
`x=1.609/0.916=1.7564`