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

1 Answer
Jul 19, 2015

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#