# Point of Intersection involving logarithms

• Potatochip911
In summary: When I tried to do it the other way it was really confusing. In summary, I attempted to solve for the point of intersection between y=log2(x) and y=log4(x), but I got lost and I'm not sure if I did something wrong.
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.

Potatochip911 said:

## 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:
$$\log_{2}{(2x)}=\log_{4}{(x)}\\ 2x=2^{\log_{4}{(x)}} \\ On the right hand side, write 2 as 41/2 . 2^{2x}=4^{\log_{4}{(x)}}\\ 2x=2\log_{4}{(x)}\\ x=\log_{4}{(x)} \\ 4^x=x \\$$
I'm really confused and I'm pretty sure I've done something wrong at this point.

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
SammyS said:

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!

Potatochip911 said:

## 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)}}=\,? \ ##

SammyS said:
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.

## What is a point of intersection involving logarithms?

A point of intersection involving logarithms is a point where two logarithmic functions intersect on a graph. This means that the two equations have the same solution.

## How do you find the point of intersection involving logarithms?

To find the point of intersection, you can set the two logarithmic equations equal to each other and solve for the variable. The resulting value is the x-coordinate of the point of intersection. You can then substitute this value into either of the equations to find the y-coordinate.

## Can there be multiple points of intersection involving logarithms?

Yes, there can be multiple points of intersection involving logarithms. This occurs when the two equations have more than one solution in common.

## What does the point of intersection represent?

The point of intersection represents the solution to the two logarithmic equations. It is the point where the two equations have the same value.

## Can a point of intersection involving logarithms be a negative number?

Yes, a point of intersection involving logarithms can be a negative number. This occurs when the solutions to the two equations are negative values that are equal to each other.

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