Please solve e^(4x+1)=7?

e^(4x+1)=7
lne^(4x+1)=ln(7)
4x+1=ln(7)
4x=ln(7)-1
....ln(7)-1
x=-----------
........4

x≈0.2364 answer//

4x + 1 = ln 7

4x = ln 7 - 1

x = ( ln 7 - 1 ) / 4

To solve this you need to insert ln into both sides of equation. You will get 4x+1=ln7
Solve for x and you will get 0.2365

x=0.2365

e^(4x + 1) = 7
4x + 1 = ln(7)
4x = ln(7) - 1
x = [ln(7) - 1]/4

e^(4x+1)=7
4x+1=ln7
4x+1=1.946
4x=1.946-1
4x=0.946
x=0.2365

You should begin by taking the natural log of both sides.
ln(e^(4x+1) = ln(7))
4x+1 = 1.95
The rest is just simple algebra
4x = 0.95
x = .2375

e^(4x+1)=7

4x+1 = ln7

4x = ln7-1

x = (ln7-1)/4

x = (1.945-1)/4

x = 0.236