# Homework Help: Point of Intersection involving logarithms

1. Jun 11, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
Solve for the point of intersection between $y=\log_{2}{(2x)}$ and $y=\log_{4}{(x)}$

2. Relevant equations
3. The attempt at a solution

Setting the two equations equal:
$$\log_{2}{(2x)}=\log_{4}{(x)}\\ 2x=2^{\log_{4}{(x)}} \\ 2^{2x}=4^{\log_{4}{(x)}}\\ 2x=2\log_{4}{(x)}\\ x=\log_{4}{(x)} \\ 4^x=x \\$$
I'm really confused and I'm pretty sure I've done something wrong at this point.

2. Jun 11, 2015

### SammyS

Staff Emeritus

If you use each side of
$\displaystyle \ 2x=2^{\log_{4}{(x)}}\$​
as the exponent with a base of 2, you get
$\displaystyle \ 2^{2x}=2^{\left(\displaystyle {2^{\log_{4}{(x)}}}\right)}\$​

Furthermore,
$\displaystyle \ \left({2^{2}}\right)^{\log_{4}{(x)}}\ne2^{\left(\displaystyle {2^{\log_{4}{(x)}}}\right)}\$​

3. Jun 11, 2015

### Potatochip911

Managed to get the correct answer after rewriting $2$ as $4^{1/2}$ like you mentioned. $$2x=2^{\log_{4}{(x)}} \\ 2^{2x}=2^{2^{\log_{4}{(x)}}}\\ 2^{2x}=2^{4^{1/2 \cdot \log_{4}{(x)}}}\\ 2^{2x}=2^{x^{1/2}} \\ 2x-\sqrt{x}=0 \\ \sqrt{x}(2\sqrt{x}-1)=0$$ Giving x=0 and x=1/4, with x=0 not being a solution since it's not in the domain, thanks for the help!

4. Jun 11, 2015

### SammyS

Staff Emeritus
A suggestion to make things a bit easier is:

Rather than using 2 as the base, use 4 as the base. Then change the base on the left hand side to 22 .

$\ \left(2^2\right)^{\log_{2}{(2x)}} =2^{\log_{2}{(4x^2)}}=\,? \$

5. Jun 11, 2015

### Potatochip911

Okay it was definitely easier to do it this way.