You need to notice that the bases 3 and 5 are primes, hence, you need to take common logarithms both sides, such that:

`lg (3^(x+1)) = lg (5^(2x-1))`

Using the power property of logarithms yields:

`(x+1)*lg 3 = (2x - 1)*lg 5`

`x*lg 3 + lg 3 = 2x*lg 5 - lg 5`

You need to isolate to one side the members that contain x, such that:

`x*lg 3 - 2x*lg 5 = -lg 3 - lg 5`

Factoring out x to the left, yields:

`x*(lg 3 - 2lg 5) = -(lg 3 + lg 5)`

`x(2lg 5 - lg 3) = lg 3 + lg 5 => x(lg 5^2 - lg 3) = lg 3 + lg 5`

Converting the difference of logarithms, to the left, into the logarithm of quotient, yields:

`x*lg(25/3) = lg 3 + lg 5`

Converting the sum of logarithms, to the right, into the logarithm of the product, yields:

`x*lg(25/3) =lg(3*5) => x = (lg 15)/(lg(25/3))`

**Hence, evaluating the solution to the given exponential equation, using common logarithms, yields **`x = (lg 15)/(lg(25/3)).`