## marcelie one year ago help please !!!! i attached the picture below

1. marcelie

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2. marcelie

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3. jojokiw3

ooo, so logs are like exponents, when you multiply exponents, what do you do with them actually?

4. marcelie

yea

5. jojokiw3

What's the answer to this question? x^2(x^2)?

6. marcelie

the answer in my sheet says x cant equal -3 and x = 2

7. jojokiw3

Hm, I don't think you can get an numerical answer to this question, because it doesn't have an equal sign.

8. jojokiw3

But you can simplify the equation more

9. marcelie

yeh one sec ..

10. freckles

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11. marcelie

12. freckles

$\log_2(x+2)+\log_2(x-1)=2?$

13. marcelie

yes.

14. jojokiw3

ooo, that makes more sense.

15. freckles

that first step you have there looks great little confused on your third line you have you should do 2^ on both sides for your second line

16. marcelie

the other one should be log base 2 (2) ?

17. marcelie

k

18. freckles

$2^{\log_2((x-1)(x+2))}=2^2$ see we do 2^ on both sides

19. freckles

this gives us: $(x-1)(x+2)=4$

20. jojokiw3

Oh, I'm sorry.

21. freckles

you will need to expand and get everything on one side like you are going to want 0 on that right hand side there

22. marcelie

hmmm. im confused

23. freckles

on what I did?

24. freckles

or what I'm asking you to do?

25. marcelie

yes. why 2 on both sides

26. freckles

$a^{\log_a(y)}=y \\ \text{ for } y>0 \text{ and } a \in (0,1) \cup (1,\infty)$

27. freckles

$\text{ that is I knew } 2^{\log_2((x+2)(x-1))}=(x+2)(x-1)$

28. freckles

for example if we had: $\log_{4}((x+2)(x-1))=2 \\ \text{ I would do 4^ on both sides } \\ 4^{(\log_4(x+2)(x-1))}=4^2 \\ \text{ which the whole point in doing the 4^ } \\ \text{ is written below } \\ (x+2)(x-1)=4^2 \\ \text{ we get \to write in its equivalent exponent form }$

29. freckles

do you understand that the following two are inverses: $y=a^x \text{ and } y=\log_a(x)$

30. freckles

$f(x)=a^x \text{ and } f^{-1}(x)=\log_a(x) \text{ are inverses iff } \\ f(f^{-1}(x))=x \text{ and } f^{-1}(f(x))=x$

31. freckles

that is: $a^{\log_a(x)}=x \text{ and } \log_a(a^x)=x$ there are some domain restrictions but yeah

32. freckles

$\text{ so we had } \log_2(stuff)=2 \\ \text{ and I did } \color{red}{2}^{\log_\color{green}{2}(stuff)}=\color{red}{2}^2 \\ stuff=4$

33. freckles

I did 2^ on both sides because of the green 2 there

34. freckles

if the green would have been 5 I would have did 5^ on both sides

35. marcelie

woahhh o.o

36. freckles

if that doesn't make sense to you I'm sure you have written log equations in their equivalent exponential form right?

37. freckles

$\log_a(y)=x \implies a^{x}=y \\ \log_2([(x-1)(x+2)]=2 \implies 2^{2}=(x-1)(x+2)$

38. marcelie

oh got it xD

39. marcelie

thank you :)

40. freckles

np