Find the solution of the exponential equation e^(2x+1)=36 in terms of logarithms, or correct to four decimal places.

e^(2x + 1) = 36       take the ln of both sides

ln e ^(2x + 1) = ln 36    and by a log property, we can write

(2x + 1) ln e  = ln 36      but ln e just = 1  ... so we have....

2x + 1 = ln 36

2x = ln 36 - 1

x = [ ln 36 - 1 ] / 2 ≈ 1.2918  {rounded}