Can anyone explain how to solve for X in the following equation 4e raised to 3x plus 1 equals 10?

Let's try natural logs.

4e^(3X+1) = 10

divide both sides by 4

e^(3X+1) = 5/2

ln[e^(3X+1)] = ln(5/2)

3X+1 lne = ln(5/2)

3X+1 = ln(5/2)/lne

( natural log of e = 1, of course )

3X + 1 = 5/2

subtract 1 from each side

3X = 3/2

divide each side by 3

X = 1/2