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

1 Answer
Apr 7, 2018

Answer:

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!