Thread: similar logarithm problem!

1. similar logarithm problem!

e^2x-7e^x+10

bit confused on how to find the x?

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

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

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?

3. 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).

4. Thanks Jameson, my solution only assumes the expression is equal to zero.

5. 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!

6. 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.