# Thread: I have a Few Log related questions.

1. ## I have a Few Log related questions.

Well our professor left us quite a bit of homework over Spring Break.
And let just say he hasnt been doing the best job teaching us in class.
(me and alot of the other students agree)

Any way i need help figuring out how to solve the following problems:
20/6-e^2x=4

and

logbase4(4x)-logbase4(x/4)=2

Me and a few classmates have been breaking our heads trying to figure this out. Thank you

-Gus

2. Is the first question $\displaystyle \frac{20}{6-e^{2x}}=4$ ?

If so,

$\displaystyle 5=6-e^{2x}$

$\displaystyle e^{2x}=1$

$\displaystyle 2x=ln(1)$

$\displaystyle 2x=0$

$\displaystyle x=0$

3. Originally Posted by Danktoker
Well our professor left us quite a bit of homework over Spring Break.
And let just say he hasnt been doing the best job teaching us in class.
(me and alot of the other students agree)

Any way i need help figuring out how to solve the following problems:
20/6-e^2x=4

and

logbase4(4x)-logbase4(x/4)=2

Me and a few classmates have been breaking our heads trying to figure this out. Thank you

-Gus
sorry.. misread the question!

4. For the second one:

$\displaystyle log_4(4x)-log_4(\frac{x}{4})=log_4(16)$

$\displaystyle log_4(4x\div\frac{x}{4})=log_4(16)$

$\displaystyle log_4(\frac{16x}{x})=log_4(16)$

$\displaystyle \therefore x$ can be any real number.

5. Originally Posted by Stroodle
For the second one:

$\displaystyle log_4(4x)-log_4(\frac{x}{4})=log_4(16)$

$\displaystyle log_4(4x\div\frac{x}{4})=log_4(16)$

$\displaystyle log_4(\frac{16x}{x})=log_4(16)$

$\displaystyle \therefore x$ can be any positive real number.
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