Logarithm coefficients

[Timmeh]

Hello ,
This is my first post on the forum .
I've got an exam coming up tomorrow and I'm asking for some help to clear some stuff up .

First problem :

log₂y = 3log₂((x-2)/-2) - 2log((x²-4)/2)

question is ofcource to find the expression for y.

Now I know how to solve logarithms. Basically the coefficients 3 and 2 go as powers for the whole expression inside the logarithm so 2log₂x = log₂x² etc
Also there is a use of identity log_ax - log_ay = log_a(x/y)
Now with that I get an answer of

(x^3 - 6x^2 +12x -8) / (-2x^4 + 16x^2 - 16)

and the answer is supposed to be (2-x) / (x+2)^2

Am I doing something wrong??

Thanks in advance for the help

 

[srmichael]

Hello ,
This is my first post on the forum .
I've got an exam coming up tomorrow and I'm asking for some help to clear some stuff up .

First problem :

log₂y = 3log₂((x-2)/-2) - 2log((x²-4)/2)

question is ofcource to find the expression for y.

Now I know how to solve logarithms. Basically the coefficients 3 and 2 go as powers for the whole expression inside the logarithm so 2log₂x = log₂x² etc
Also there is a use of identity log_ax - log_ay = log_a(x/y)
Now with that I get an answer of

(x^3 - 6x^2 +12x -8) / (-2x^4 + 16x^2 - 16)

and the answer is supposed to be (2-x) / (x+2)^2

Am I doing something wrong??

Thanks in advance for the help

I assume the second log on the right of the equal sign is supposed to have a base of 2 as well.

We can't determine your mistake unless we see your work. Also, I think the answer is missing a 2 in the denominator.

That being said:

\(\displaystyle log_{2}y=3log_{2}(\dfrac{x-2}{-2})-2log_{2}(\dfrac{x^2-4}{2})\)

\(\displaystyle log_{2}y=log_{2}(\dfrac{x-2}{-2})^3-log_{2}(\dfrac{x^2-4}{2})^2\)

\(\displaystyle log_{2}y=log_{2}\dfrac{(\dfrac{x-2}{-2})^3}{(\dfrac{x^2-4}{2})^2}\)

\(\displaystyle y=\dfrac{(\dfrac{x-2}{-2})^3}{(\dfrac{x^2-4}{2})^2}\)

\(\displaystyle y=\dfrac{\dfrac{(x-2)^3}{-8}}{\dfrac{(x^2-4)^2}{4}}\)

\(\displaystyle y=\dfrac{(x-2)^3}{-8} \cdot \dfrac{4}{(x^2-4)^2}\)

\(\displaystyle y=\dfrac{(x-2)^3}{-8} \cdot \dfrac{4}{[(x+2)(x-2)]^2}\)

\(\displaystyle y=\dfrac{(x-2)^3}{-8} \cdot \dfrac{4}{(x+2)^2(x-2)^2}\)

\(\displaystyle y=-\dfrac{x-2}{2} \cdot \dfrac{1}{(x+2)^2}\)

\(\displaystyle y=-\dfrac{x-2}{2(x+2)^2}\)

\(\displaystyle y=\dfrac{2-x}{2(x+2)^2}\)

 

[Timmeh]

Thanks alot!

I did it exactly how you did , the only trick was not to multiply the (x-2)^3 like I did but wait for it to cancel out .
I knew there was some trick that I was missing .
Once again , thanks