# Math Help - Logs Help

1. ## Logs Help

hi all. i've been stuck on this logarithm question for a while now, and i just can't seem to get the answer.

the question is
$e^(4x^2 - 2) = 3$
(the 4x^2 - 2 is supposed to be the power of e)
solve for x to 2 decimals.

anyways, i keep on getting a positive and negative number answer, but it's the incorrect value. i get like +- 0.88 , so i'm just curious as to how i can solve this question.

thank you very much

2. Originally Posted by mneox
hi all. i've been stuck on this logarithm question for a while now, and i just can't seem to get the answer.

the question is
$e^(4x^2 - 2) = 3$
(the 4x^2 - 2 is supposed to be the power of e)
solve for x to 2 decimals.

anyways, i keep on getting a positive and negative number answer, but it's the incorrect value. i get like +- 0.88 , so i'm just curious as to how i can solve this question.

thank you very much
your values are correct ... why do you think they are wrong?

3. Originally Posted by mneox
hi all. i've been stuck on this logarithm question for a while now, and i just can't seem to get the answer.

the question is
$e^(4x^2 - 2) = 3$
(the 4x^2 - 2 is supposed to be the power of e)
solve for x to 2 decimals.

anyways, i keep on getting a positive and negative number answer, but it's the incorrect value. i get like +- 0.88 , so i'm just curious as to how i can solve this question.

thank you very much
$e^{4x^2 - 2} = 3$

$4x^2 - 2 = \ln{3}$

$4x^2 = 2 + \ln{3}$

$x^2 = \frac{1}{2} + \frac{1}{4}\ln{3}$

$x = \pm\sqrt{\frac{1}{2} + \frac{1}{4}\ln{3}}$.

Therefore

$x = -\sqrt{\frac{1}{2} + \frac{1}{4}\ln{3}}$ or $x = \sqrt{\frac{1}{2} + \frac{1}{4}\ln{3}}$.

Now put these into your calculator for the decimal approximations.

4. Originally Posted by Prove It
$e^{4x^2 - 2} = 3$

$4x^2 - 2 = \ln{3}$

$4x^2 = 2 + \ln{3}$

$x^2 = \frac{1}{2} + \frac{1}{4}\ln{3}$

$x = \pm\left(\frac{1}{2} + \frac{1}{4}\ln{3}\right)$.

forget a square root?

Therefore

$x = -\frac{1}{2} - \frac{1}{4}\ln{3}$ or $x = \frac{1}{2} + \frac{1}{4}\ln{3}$.

Now put these into your calculator for the decimal approximations.
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5. Originally Posted by skeeter
your values are correct ... why do you think they are wrong?
for one, we have a answer key and it says that the correct value is +- 1.76.

but if what i got was correct, maybe it's just the answer key is incorrect?

Originally Posted by Prove It
$e^{4x^2 - 2} = 3$

$4x^2 - 2 = \ln{3}$

$4x^2 = 2 + \ln{3}$

$x^2 = \frac{1}{2} + \frac{1}{4}\ln{3}$

$x = \pm\left(\frac{1}{2} + \frac{1}{4}\ln{3}\right)$.

Therefore

$x = -\frac{1}{2} - \frac{1}{4}\ln{3}$ or $x = \frac{1}{2} + \frac{1}{4}\ln{3}$.

Now put these into your calculator for the decimal approximations.

i was just wondering how you got the same answer but with a +/- after you square root?

$x^2 = \frac{1}{2} + \frac{1}{4}\ln{3}$

$x = \pm\left(\frac{1}{2} + \frac{1}{4}\ln{3}\right)$

6. try your solutions in the original equation , then try the "book" answers.

see for yourself which are correct.

7. Originally Posted by skeeter
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Yes I did - edited.