anonymous
  • anonymous
Solve the logarithmic equation for x: log base 2(x+1)+log base 2(x-1)=3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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nowhereman
  • nowhereman
Apply exponentiation!
anonymous
  • anonymous
.........huh? :P
nowhereman
  • nowhereman
If you know the logarithm you should know the exponential.

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anonymous
  • anonymous
I'm lost........ Gah why does math have to be so hard?
nowhereman
  • nowhereman
It's not, if you follow the definitions. How did you define log?
anonymous
  • anonymous
What do you mean? I didn't define anything... I don't think...
nowhereman
  • 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
  • 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
  • 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
  • anonymous
Is b the base?
nowhereman
  • nowhereman
Indeed
anonymous
  • anonymous
So it would be \[2^\log2(3)=3\]?
watchmath
  • 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
  • anonymous
That's what I thought, cause I didn't remember anything like that.... Help?
watchmath
  • watchmath
Do you remember this property? \(log_b A+\log_b B= \log_b(AB)?\)
anonymous
  • anonymous
I think so...
watchmath
  • watchmath
Ok, what do you get if you apply that property to the left hand side of your equation?
anonymous
  • anonymous
\[\log_{2} 2x\]I think. Because the +1 and the -1 would cancel out.
watchmath
  • watchmath
But you multiply the inside instead of adding them.
anonymous
  • anonymous
Why would I multiply them? It says I'm adding the two logs...
anonymous
  • anonymous
Answer x = 3; log(3+1) to the base 2 = 2 and log(3-1) to the base 2 = 1
watchmath
  • watchmath
look at again \(log_b A +\log_b B=\log_b(AB)\) In there you multiply A and B
anonymous
  • anonymous
Ohhhhh okay. So it would be\[\log_{2} x^2+1\]
watchmath
  • watchmath
Check it again what do you get if you do (x+1)(x-1) (foil it :) )
anonymous
  • 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
  • anonymous
\[\log_{2} x^2-1\]
anonymous
  • anonymous
that would equal 3, right?
watchmath
  • 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
  • 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
  • watchmath
Remember this \[\log_bx=y\iff b^y=x\]
anonymous
  • anonymous
That's right
anonymous
  • anonymous
OHH I REMEMBER THAT! :D
anonymous
  • anonymous
So it would be \[2^3=x^2-1\]
anonymous
  • anonymous
Yep. Now solve for x :)
anonymous
  • anonymous
\[x^2=9\]so x=3 :D
anonymous
  • anonymous
AHHH YOU GUYS ARE AMAZING<3
anonymous
  • anonymous
x could be -3 as well. Since we are dealing with logarithms, we ignore -3
anonymous
  • anonymous
Oh yeah, I forgot about that. Negatives don't work with logs.
watchmath
  • 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
  • anonymous
YAYAYAYYAYAYAYAYYAYAY!<3 Thanks so much guys!
anonymous
  • anonymous
Any time!

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