## anonymous one year ago Evalute log 2(x-2) + log 2 (x+5)=3

1. anonymous

@Zarkon

2. Nnesha

familiar with the log properties ?

3. anonymous

yes

4. Nnesha

okay so apply one of them at left side $\huge\rm \log_2(x-2) \color{reD}{+} \log_2 (x+5)=3$ there is plus sign whichnoe one would you use ? quotient or product property ?

5. anonymous

product

6. Nnesha

yes $\huge\rm log_b x + \log_b y = \log_b(x \times y)$ try to apply this on ur original question pretty sure you can do it!

7. anonymous

x^2+3x-10

8. Nnesha

yes $\huge\rm log_2 (x^2+3x-10)=3$ now we have to convert log to exponential form

9. anonymous

x^2+3x=13

10. Nnesha

how did you get 13 ?

11. Nnesha

ohh well log_2 is still there $\huge\rm log_b x + \log_b y = \color{ReD}{\log_b}(x \times y)$ $\huge\rm\color{reD}{ log_2} (x^2+3x-10)=3$ log_b is a common factor

12. anonymous

now what do i do?

13. UnkleRhaukus

now, exponentiate both sides of the equation, with the base 2

14. anonymous

What does that mean?

15. Nnesha

|dw:1439613747571:dw| convert log to exponential form

16. UnkleRhaukus

from$LHS = RHS$ $2^{LHS} = 2^{RHS}$

17. anonymous

ok hold on let me try..

18. anonymous

y=2^x ?? Im so confused.

19. UnkleRhaukus

\begin{align}\log_2(x^2+3x-10)&=3\\ 2^{\log_2(x^2+3x-10)}&=2^3\\ (x^2+3x-10)&=2^3\end{align}

20. Nnesha

that was an example |dw:1439613949291:dw|

21. anonymous

Yay! :) Would that be the final answer?

22. UnkleRhaukus

23. anonymous

I can do that :) Please hold on

24. anonymous

(x-5)(x+2)

25. anonymous

wait! I factored wrong! Ughhh hold again

26. anonymous

(x+5)(x-2)

27. UnkleRhaukus

\begin{align}\log_2(x^2+3x-10)&=3\\ 2^{\log_2(x^2+3x-10)}&=2^3\\ (x^2+3x-10)&=2^3\\ x^2+3x-18&=0\\ &= \end{align}

28. anonymous

(x+6)(x-3)

29. UnkleRhaukus

$x^2+3x-18=0\\ (x+6)(x-3)=0$ good, so what values of $$x$$ do you get?

30. anonymous

x=-6,3

31. UnkleRhaukus

If we try plugging the first of these ($$x=-6$$) back into the original equation $\log _2(x-2) +\log_2 (x+5)=3$ we get$\log _2(-6-2) +\log_2 (-6+5)=3\\\log _2(-8) +\log_2 (-1)=3$ .... logs of negative numbers... hmm doesn't make much sense. If we plugging the second of these ($$x=3$$) back into the original equation $\log _2(x-2) +\log_2 (x+5)=3$ we get$\log _2(3-2) +\log_2 (3+5)=3\\\log _2(1) +\log_2 (8)=3\\0+\log_2 (2^3)=3\\3\log_2 (2)=3 \\3=3$ which is always true. so this is our solution

32. anonymous

Thanks :)!! I understand it now!!!

33. anonymous

If I post another question, will u answer it?

34. anonymous

solve 7-7x=(3x+2)(x-1) and check for extraneous solutions

35. anonymous

7 - 7x = 3x² - x - 2 3x² + 6x - 9 = 0 3(x + 3)(x - 1) = 0 x = -3, 1

36. anonymous

Would it be extraneous? Thanks geny55!!! :D

37. anonymous