Question

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

Answer 1

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`