Question

How do you solve `2* 3^x = 7* 5^x`?

Answer 1

I found: `x=-7.0421`

Explanation:

Try rearranging it: as:
`3^x/5^x=7/2`
write it as:
`(3/5)^x=7/2`
use the definition of logarithm to write:
`log_(3/5)(7/2)=x`
We can now use the change of base formula to transform our log into a quocient of natural logs (these can be evaluated using a pocket calculator) as:
`x=(ln(7/2))/(ln(3/5))=-7.0421`