Once you got to the b^x=y form, you did not need to do all of that work and instead should have taken the log to the base that matches whatever the exponential base is. In b^x=y, you would apply log to base b on both sides of the equation, canceling the log and exponential of the same base, meaning x = log of y to base b. It would have been fine to leave your response in log form. When using logs to solve exponential equations, you should use minimal steps assuming the exponential equation is set up to handle such cases.
Very nice
Once you got to the b^x=y form, you did not need to do all of that work and instead should have taken the log to the base that matches whatever the exponential base is. In b^x=y, you would apply log to base b on both sides of the equation, canceling the log and exponential of the same base, meaning x = log of y to base b. It would have been fine to leave your response in log form. When using logs to solve exponential equations, you should use minimal steps assuming the exponential equation is set up to handle such cases.
Once you had a numerical fraction, just solve for X by doing the simple division. I don't understand why you made the problem more complicated.
5^(x + 1) = 2^(3x)
5^(x) * 5^(1) = [2^(3)]^(x)
5^(x) * 5 = 8^(x)
5^(x) / 8^(x) = 1/5
(5/8)^(x) = 1/5
Ln[(5/8)^(x)] = Ln(1/5)
x.Ln(5/8) = - Ln(5)
x = - Ln(5) / Ln(5/8)
x = Ln(5) / Ln(8/5)