How do you solve #e^(2x) = ln(4 + e)#?

1 Answer
Gió
May 27, 2016

I found: #x=(ln(ln(4+e)))/2=0.322197#

Explanation:

We could call the constant #ln(4+e)#, say, #k# and write:
#e(^2x)=k#
take the natural log of both sides:
#ln(e^(2x))=ln(k)#
and:
#2x=ln(k)# where #ln# and #e# eleiminated one another;
finally:
#x=ln(k)/2#
Now:
if we can use a calculator we have that:
#k=ln(4+e)=1.90483#
so that #x# becomes:
#x=(ln(ln(4+e)))/2=0.322197#

Related questions
See all questions in Logarithmic Models
Impact of this question
2287 views around the world
You can reuse this answer
Creative Commons License