Question

How do you solve `4(2^x) = 5^x`?

Answer 1

I found: `x=1.5129`

Explanation:

Start by writing it as:
`2^2*2^x=5^x`
`2^(x+2)=5^x`
take the `ln` of both sides:
`ln(2^(x+2))=ln5^x`
use the property of logs:
`logx^a=alogx`
to get:
`(x+2)ln2=xln5`
`xln2+2ln2-xln5=0`
isolate `x`:
`x(ln2-ln5)=-2ln2`
`x=-(2ln2)/(ln2-ln5)=-1.3863/-0.9163=1.5129`