Question

need help. solve for x.
\[e ^{x ^{2}-1}=1/x\]

Answer 1

Are you allowed to use Logarithms??

Answer 2

Taking ln both the sides:

\[\ln_e(e^{x^2-1}) = \ln_e(\frac{1}{x})\]

\[\ln_e(e^{x^2-1}) = \ln_e(1) - \ln_e(x)\]

\[(x^2 - 1)(1) = 0 - \ln_e(x)\]

\[1 - x^2 = \ln_e(x)\]

Got stuck now.

Answer 3

we know that x>0
for x>1 \( e^{x^2-1}>1 \ \ and \ \ \frac{1}{x}<1 \)

for x<1 \( e^{x^2-1}<1 \ \ and \ \ \frac{1}{x}>1 \)

so x=1

Answer 4

yeah i think my teacher would like us to use logs

Answer 5

@waterineyes
use the same approach for ur solution

Answer 6

Yeah for x = 1,

\(1- (1)^2 = ln_e(1)\)

\(0 = 0\)

Yeah x = 1 can be one possibility..

Answer 7

And I think there is one possibility only...

So,

\[\large \color{green}{x = 1}\]

Answer 8

ok thank you for your help!

Answer 9

Welcome dear..

Answer 10

Say thanks to mukushla also @jkd13

Answer 11

it was a thanks to both of you :)

Answer 12

the method used here is the method i'd use but there's a problem... you'll end up with like what waterineye got and that is an \(\large x^2 \) term and a \(\large lnx \) term. the problem here is to isolate the x and there is no elematary algebraic method that i know of to do that.....

Answer 13

ok thank you cause thats whats throwing me off but i think this will be one to ask my instructor to see what he's looking for.

Answer 14

yw...:)