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

1. nowhereman

Apply exponentiation!

2. anonymous

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

3. nowhereman

If you know the logarithm you should know the exponential.

4. anonymous

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

5. nowhereman

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

6. anonymous

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

7. 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?

8. 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.

9. 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.

10. anonymous

Is b the base?

11. nowhereman

Indeed

12. anonymous

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

13. 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.

14. anonymous

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

15. watchmath

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

16. anonymous

I think so...

17. watchmath

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

18. anonymous

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

19. watchmath

20. anonymous

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

21. anonymous

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

22. watchmath

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

23. anonymous

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

24. watchmath

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

25. 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

26. anonymous

$\log_{2} x^2-1$

27. anonymous

that would equal 3, right?

28. watchmath

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

29. 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...

30. watchmath

Remember this $\log_bx=y\iff b^y=x$

31. anonymous

That's right

32. anonymous

OHH I REMEMBER THAT! :D

33. anonymous

So it would be $2^3=x^2-1$

34. anonymous

Yep. Now solve for x :)

35. anonymous

$x^2=9$so x=3 :D

36. anonymous

AHHH YOU GUYS ARE AMAZING<3

37. anonymous

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

38. anonymous

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

39. 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.

40. anonymous

YAYAYAYYAYAYAYAYYAYAY!<3 Thanks so much guys!

41. anonymous

Any time!