Point of Intersection involving logarithms

[Potatochip911]

Homework Statement

Solve for the point of intersection between $y=\log_{2}{(2x)}$ and $y=\log_{4}{(x)}$

Homework Equations

3. The Attempt at a Solution [/B]
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.

 

[SammyS]

Homework Statement

Solve for the point of intersection between $y=\log_{2}{(2x)}$ and $y=\log_{4}{(x)}$

Homework Equations

3. The Attempt at a Solution [/B]
Setting the two equations equal:
$$\log_{2}{(2x)}=\log_{4}{(x)}\\
2x=2^{\log_{4}{(x)}} \\ On the right hand side, write 2 as 4^1/2 .
2^{2x}=4^{\log_{4}{(x)}}\\
2x=2\log_{4}{(x)}\\
x=\log_{4}{(x)} \\
4^x=x \\ $$
I'm really confused and I'm pretty sure I've done something wrong at this point.

You're right about that.

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)}\$​

 

[Potatochip911]

You're right about that.

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)}\$​

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!

 

[SammyS]

Homework Statement

Solve for the point of intersection between $y=\log_{2}{(2x)}$ and $y=\log_{4}{(x)}$

Homework Equations

3. The Attempt at a Solution [/B]
Setting the two equations equal:
$$\log_{2}{(2x)}=\log_{4}{(x)}\\
2x=2^{\log_{4}{(x)}}$$

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)}}=\,? \$

 

[Potatochip911]

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)}}=\,? \$

Okay it was definitely easier to do it this way.