Solve the logarithmic equation for x: log base 2(x+1)+log base 2(x-1)=3

- anonymous

Solve the logarithmic equation for x: log base 2(x+1)+log base 2(x-1)=3

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- nowhereman

Apply exponentiation!

- anonymous

.........huh? :P

- nowhereman

If you know the logarithm you should know the exponential.

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## More answers

- anonymous

I'm lost........ Gah why does math have to be so hard?

- nowhereman

It's not, if you follow the definitions. How did you define log?

- anonymous

What do you mean? I didn't define anything... I don't think...

- nowhereman

I mean in class, or where you learned about the logarithm. Or how do you want to solve any problem containing it, if you don't know what it is?

- anonymous

I learned this like 6 months ago, so I really don't remember. This is a take home test... I just need to solve for x, I think.

- nowhereman

Then you should look it up, in your notes actually, but wikipedia is often good too.
What you need here is \[b^{\log_b x} = x\] because logarithm is the inverse function of the exponential.

- anonymous

Is b the base?

- nowhereman

Indeed

- anonymous

So it would be \[2^\log2(3)=3\]?

- watchmath

hi nowhere man, we need to combine the logs first before applying exponentiation. Also in US usually they don't teach exponentiation explicitly in that way.

- anonymous

That's what I thought, cause I didn't remember anything like that.... Help?

- watchmath

Do you remember this property?
\(log_b A+\log_b B= \log_b(AB)?\)

- anonymous

I think so...

- watchmath

Ok, what do you get if you apply that property to the left hand side of your equation?

- anonymous

\[\log_{2} 2x\]I think. Because the +1 and the -1 would cancel out.

- watchmath

But you multiply the inside instead of adding them.

- anonymous

Why would I multiply them? It says I'm adding the two logs...

- anonymous

Answer x = 3;
log(3+1) to the base 2 = 2
and log(3-1) to the base 2 = 1

- watchmath

look at again \(log_b A +\log_b B=\log_b(AB)\)
In there you multiply A and B

- anonymous

Ohhhhh okay. So it would be\[\log_{2} x^2+1\]

- watchmath

Check it again what do you get if you do (x+1)(x-1) (foil it :) )

- anonymous

Formula
log(A) to the base 2 + log(B) to the base 2 = log(A B) to the base 2;
Steps:
1. log (x+1)(x-1) to the base 2 = 3
2. removing log, 2 raised to 3 = (x+1) (x-1)
3. 8 = x^2 - 1
4. 9 = x^2
5. taking square root on both sides, 3 = x

- anonymous

\[\log_{2} x^2-1\]

- anonymous

that would equal 3, right?

- watchmath

great!
So now we have
\(\log_2(x^2-1)=3\)
Do you remember how to change from log equation into exponential equation?

- anonymous

I think if I make the three on the other side a log as well I could do it, but that doesn't seem right...

- watchmath

Remember this
\[\log_bx=y\iff b^y=x\]

- anonymous

That's right

- anonymous

OHH I REMEMBER THAT! :D

- anonymous

So it would be \[2^3=x^2-1\]

- anonymous

Yep. Now solve for x :)

- anonymous

\[x^2=9\]so x=3 :D

- anonymous

AHHH YOU GUYS ARE AMAZING<3

- anonymous

x could be -3 as well. Since we are dealing with logarithms, we ignore -3

- anonymous

Oh yeah, I forgot about that. Negatives don't work with logs.

- watchmath

To bee a little pedantic from \(x^2=9\) we have \(x=\pm 3\). But when we plug in back x=-3 into the equation we are in trouble because we can't compute log of negative number. SO x=3 is the only solution.

- anonymous

YAYAYAYYAYAYAYAYYAYAY!<3 Thanks so much guys!

- anonymous

Any time!

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