Question

can anyone explain the step in going from

ln(x+6)=3

to
e^3=x+6

Answer 1

\[e ^{\ln a} = a\]That is the key to this problem. Also, if you ask yourself, "what is the ln a?", you will come to realize that ln a is the exponent that you put on "e" ot get "a".

So, you just use each side as an exponent on e. Raise both sides as a power of "e".

Answer 2

So, if you let a = x + 6, you have:
\[e ^{\ln (x + 6)} = x + 6\]