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anonymous
 5 years ago
Solve the logarithmic equation for x: log base 2(x+1)+log base 2(x1)=3
anonymous
 5 years ago
Solve the logarithmic equation for x: log base 2(x+1)+log base 2(x1)=3

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nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1Apply exponentiation!

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1If you know the logarithm you should know the exponential.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm lost........ Gah why does math have to be so hard?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1It's not, if you follow the definitions. How did you define log?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you mean? I didn't define anything... I don't think...

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1I 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.1Then 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
 5 years ago
Best ResponseYou've already chosen the best response.0So it would be \[2^\log2(3)=3\]?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0hi 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
 5 years ago
Best ResponseYou've already chosen the best response.0That's what I thought, cause I didn't remember anything like that.... Help?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Do you remember this property? \(log_b A+\log_b B= \log_b(AB)?\)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, what do you get if you apply that property to the left hand side of your equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\log_{2} 2x\]I think. Because the +1 and the 1 would cancel out.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0But you multiply the inside instead of adding them.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why would I multiply them? It says I'm adding the two logs...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Answer x = 3; log(3+1) to the base 2 = 2 and log(31) to the base 2 = 1

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0look at again \(log_b A +\log_b B=\log_b(AB)\) In there you multiply A and B

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ohhhhh okay. So it would be\[\log_{2} x^2+1\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Check it again what do you get if you do (x+1)(x1) (foil it :) )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Formula log(A) to the base 2 + log(B) to the base 2 = log(A B) to the base 2; Steps: 1. log (x+1)(x1) to the base 2 = 3 2. removing log, 2 raised to 3 = (x+1) (x1) 3. 8 = x^2  1 4. 9 = x^2 5. taking square root on both sides, 3 = x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that would equal 3, right?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0great! So now we have \(\log_2(x^21)=3\) Do you remember how to change from log equation into exponential equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0Remember this \[\log_bx=y\iff b^y=x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OHH I REMEMBER THAT! :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So it would be \[2^3=x^21\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep. Now solve for x :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0AHHH YOU GUYS ARE AMAZING<3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x could be 3 as well. Since we are dealing with logarithms, we ignore 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh yeah, I forgot about that. Negatives don't work with logs.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0To 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
 5 years ago
Best ResponseYou've already chosen the best response.0YAYAYAYYAYAYAYAYYAYAY!<3 Thanks so much guys!
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