e^2x-7e^x+10

bit confused on how to find the x?

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- Oct 2nd 2009, 02:07 PMfrozenflamessimilar logarithm problem!
e^2x-7e^x+10

bit confused on how to find the x? - Oct 2nd 2009, 02:53 PMJameson
- Oct 2nd 2009, 02:53 PMpickslides
$\displaystyle e^{2x}-7e^x+10=0$

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

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

- Oct 2nd 2009, 02:59 PMfrozenflames
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 PMJameson