2(5)^x = 3^(x + 1). Solve for x

Solve for x:

2(5)^x = 3^(x + 1)

I dont know how to make the bases the same in order to solve for the x in the exponent.

Solve for x:
7^(3x + 1) = 5^x

2(5)^x = 3^(x + 1)
2*(5)^x = 3*3^(x )
(5)^x / (3)^x = 3/2
(5/3)^x = 3/2
x = Log(3/2) / Log(5/3) = (Log(3)-Log(2))/(Log(5)-Log(3))

7^(3x + 1) = 5^x
7 * 7^(3x) = 5^x
7 * (7^3)^x = 5^x
((7^3)/5)^x = 1/7
x = Log(1/7)/Log((7^3)/5) = (-Log(7))/(3Log(7) - Log(5))