Question

Solve for x: e^2x=3x^2

Answer 1

\[e^{2x}=3x^2\]?

Answer 2

yes

Answer 3

im not sure we can.

Answer 4

using a graphing calculator?

Answer 5

i could be wrong:) looks like mathstudent has something

Answer 6

well you can get an approximate in a calculator
solve(e^(2x)=3x^2,x)
this is how you would do it on a ti-89

Answer 7

I have a ti-84

Answer 8

same thing should work

Answer 9

lets see what @mathstudent55 says

Answer 10

okay!

Answer 11

http://www.wolframalpha.com/input/?i=e%5E%282x%29+%3D+3x%5E2

Answer 12

\(e^{2x} = 3x^2 \)

\(\ln (e^{2x}) = \ln(3x^2) \)

\(2x = \ln3 + 2 \ln x \)

\( 2x - 2 \ln x = \ln 3\)

\( 2(x - \ln x) = \ln 3 \)

\(x - \ln x = \dfrac{\ln 3}{2} \)

I can't go any further.

Answer 13

I think its transcendental and cant be solved.
for an approximate
http://www.wolframalpha.com/input/?i=approximate+e%5E%282x%29+%3D+3x%5E2

Answer 14

http://en.wikipedia.org/wiki/Transcendental_equation

read that @lily104

Answer 15

thank you!

Answer 16

\[e^{u}=\sum_0\frac1{n!}u^n\]
\[e^{2x}=\sum_0\frac1{n!}(2x)^n\]
\[e^{2x}=\sum_0\frac{2^n}{n!}x^n\]
\[e^{2x}=1+\sum_1\frac{2^n}{n!}x^n\]
\[e^{2x}=1+2x+\sum_2\frac{2^n}{n!}x^n\]
\[\color{red}{3x^2} =1+2x+2x^2+\sum_3\frac{2^n}{n!}x^n\]
\[0=1+2x+(2x^2\color{red}{-3x^2})+\sum_3\frac{2^n}{n!}x^n\]
\[0=1+2x-x^2+\sum_3\frac{2^n}{n!}x^n\]
\[x^2-2x+1=\sum_3\frac{2^n}{n!}x^n\]:)

Answer 17

.. x^2-2x-1 that is