Question

How do you solve for x in `5^x=4^(x+1)`?

Answer 1

`xapprox6.21`

Explanation:

First we'll take the `log` of both sides:
`log(5^x)=log(4^(x+1))`
Now there's a rule in logarithms which is: `log(a^b)=blog(a)`, saying that you can move any exponents down and out of the `log` sign. Applying this:
`xlog5=(x+1)log4`
Now just rearrange to get x on one side
`xlog5=xlog4+log4`
`xlog5-xlog4=log4`
`x(log5-log4)=log4`
`x=log4/(log5-log4)`
And if you type that into your calculator you'll get:
`xapprox6.21...`