Question

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

Answer 1

Apply exponentiation!

Answer 2

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

Answer 3

If you know the logarithm you should know the exponential.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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?

Answer 8

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.

Answer 9

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.

Answer 10

Is b the base?

Answer 11

Indeed

Answer 12

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

Answer 13

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.

Answer 14

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

Answer 15

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

Answer 16

I think so...

Answer 17

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

Answer 18

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

Answer 19

But you multiply the inside instead of adding them.

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

that would equal 3, right?

Answer 28

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

Answer 29

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...

Answer 30

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

Answer 31

That's right

Answer 32

OHH I REMEMBER THAT! :D

Answer 33

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

Answer 34

Yep. Now solve for x :)

Answer 35

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

Answer 36

AHHH YOU GUYS ARE AMAZING<3

Answer 37

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

Answer 38

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

Answer 39

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.

Answer 40

YAYAYAYYAYAYAYAYYAYAY!<3 Thanks so much guys!

Answer 41

Any time!