How to solve ln(x)=2x I know this question might be an easy one. but it has been so long since I solved such questions and I didn't find a an explanation on the internet. I'd like if someone can remind me. I reached that e^(2x) = x , but didn't know how to continue from here. I remember something that has to do with bases and equalizing parameters, but how do I do that in this case?

Aron Heath

Aron Heath

Answered question

2022-11-11

How to solve ln ( x ) = 2 x
I know this question might be an easy one. but it has been so long since I solved such questions and I didn't find a an explanation on the internet. I'd like if someone can remind me.
I reached that e 2 x = x, but didn't know how to continue from here. I remember something that has to do with bases and equalizing parameters, but how do I do that in this case?

Answer & Explanation

cismadmec

cismadmec

Beginner2022-11-12Added 22 answers

Draw a graph. log x < 2 x
A proof is by noting that log x < 2 x for x < 1 and then differentiating both sides to see that the LHS grows slower than the RHS.
Equivalently, e 2 x > x
spasiocuo43

spasiocuo43

Beginner2022-11-13Added 6 answers

Since e x x + 1, we have
e 2 x ( x + 1 ) 2 = x 2 + 2 x + 1 = ( x + 1 2 ) 2 + 3 4 + x > x
so there are no solutions (because they can never be equal).
I like using e x x + 1 rather than calculus.

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