Question

How do you solve `2^ { x } \cdot 3^ { 2x - 1} = 5^ { x + 1}`?

Answer 1

`x==2.1141`

Explanation:

Taking logarithm (base10) on both sides of `2^x*3^(2x-1)=5^(x+1)`, we get

`xlog2+(2x-1)log3=(x+1)log5`

or `x(log2+2log3-log5)=log5+log3`

or `xlog((2*3^2)/5)=log15`

or `xlog3.6=log15`

or `x=log15/log3.6=1.1761/0.5563=2.1141`