# similar logarithm problem!

• Oct 2nd 2009, 02:07 PM
frozenflames
similar logarithm problem!
e^2x-7e^x+10

bit confused on how to find the x?
• Oct 2nd 2009, 02:53 PM
Jameson
Quote:

Originally Posted by frozenflames
e^2x-7e^x+10

bit confused on how to find the x?

This can't be solved because it doesn't equal anything. Also, use parentheses to avoid confusion. You could mean e^(2)*x or e^(2x).
• Oct 2nd 2009, 02:53 PM
pickslides
$\displaystyle e^{2x}-7e^x+10=0$

$\displaystyle (e^{x})^2-7e^x+10=0$

Lets make $\displaystyle e^{x} = a$

Now we have

$\displaystyle a^2-7a+10=0$

which can be factored as follows

$\displaystyle (a-5)(a-2)=0$

by the null factor law

$\displaystyle a=2,5$

and bring our original unknown back into the picture

$\displaystyle e^x=2,5$

$\displaystyle x=\ln(2),\ln(5)$

Does this make sense?
• Oct 2nd 2009, 02:57 PM
pickslides
Thanks Jameson, my solution only assumes the expression is equal to zero.
• Oct 2nd 2009, 02:59 PM
frozenflames
okay, that makes a lot of sense so there are two answers technically and I forgot to take the ln of the 2 and 5. that is where i was going wrong..thanks so much!
• Oct 2nd 2009, 03:13 PM
Jameson
Quote:

Originally Posted by pickslides
Thanks Jameson, my solution only assumes the expression is equal to zero.

Given the way the problem worked out, it seems that was a good assumption. It's just not good to see people solve for things that can't be solved.