How do you simplify e^[log_e^2 (9)]?

1 Answer
Apr 7, 2018

The most simplified I could get it is 81^ln3 which is about 124.935.

Explanation:

Use these logarithmic properties:

color(blue)(log_e)x=color(blue)(ln)x

log^2x=(logx)^2

log(color(blue)x^color(red)a)=color(red)alogcolor(blue)x

And these exponential properties:

(x^color(red)m)^color(blue)n=x^(color(red)mcolor(blue)n)=(x^color(blue)n)^color(red)m

e^lncolor(blue)x=color(blue)x

Here's the expression:

color(white)=e^(log_e^2 9)

=e^(ln^2(9))

=e^(ln^2 9)

=e^((ln 9)^2)

=e^((ln (3^2))^2)

=e^((2ln3)^2)

=e^(2^2ln^2 3)

=e^(4ln^2 3)

=(e^(ln^2 3))^4

=(e^(ln3*ln3))^4

=((e^(ln 3))^ln3)^4

=((color(red)cancelcolor(black)e^(color(red)cancelcolor(black)cc(ln) 3))^ln3)^4

=(3^ln3)^4

=3^(4ln3)

=(3^4)^ln3

=81^ln3

~~124.93528000...

Hope this helped!