Question

How do you solve `11.3^(2x – 1) = 15.7`?

Answer 1

I found: `x=1.0678`

Explanation:

Here I would use the natural log on both sides and one property of logs (about the exponent of the argument):

I write:

`color(red)(ln)11.3^(2x-1)=color(red)(ln)15.7`

then:

`(2x-1)ln(11.3)=ln(15.7)`

rearrange:
`2x-1=(ln(15.7))/(ln(11.3))`
`2x=(ln(15.7))/(ln(11.3))+1`
`x=[(ln(15.7))/(ln(11.3))+1]/2=1.0678`