## anonymous 4 years ago Have a few problems I was stuck on. 1)log_9(x+4)+log_9(x-4)=1 2)Sqrt(3x-2)-sqrt(x)=1

1. anonymous

First combine the logs ( addition of log is multiplication) log(x+4) + log(x-4) = Log ((x+4)/(x-4)) raise everything from base 9 for part one, doing this cancels the log because they are inverse functions and 9^1 is just 9 resulting in (x+4)(x-4) = 9 distribute x^2 -16 = 9 x^2 = 25 x= + or -5

2. anonymous

typo, the log product should read log ((X+4)(x-4))

3. anonymous

ok lets solve it since $\Large \log(a)+\log(b)=\log(ab)$ (addition changes to multiplication $\Large \log _{9}((x+4)(x-4))=1$ since $\Large a^2-b^2=(a+b)(a-b)$ so $\Large \log _{9}(x^2-16)=1$

4. anonymous

Yes, I got as far as @sami-21 But got stuck. I knew the answer was 5. Didnt know how to get it.

5. anonymous

so now we can cgange the log by change of bass formula

6. anonymous

change*

7. anonymous

Hmm..thats where i got stuck. How do you do that?

8. anonymous

$\Large \log _{a}(b)=\frac{\ln(a)}{\ln{b}}$

9. anonymous

it is upto use to change it to any base .i used natural log whose base is e

10. anonymous

Hmm...i think i get it.

11. anonymous

ok

12. anonymous

I had another one similar to it: 7^(5x-2)=5^(3-x)

13. anonymous

I know you have to log both sides.

14. anonymous

yes take ln of both sides

15. anonymous

then use property $\Large \log _{a}(b)^n=nlog _{a}b$ which means exponents gets down and gets multiplied !

16. anonymous

let me do the your question now using above property

17. anonymous

take ln of both sides$\Large \ln(7)^{5x-2}=\ln(5)^{3-x}$ so using the above mentioned property $\Large (5x-2)\ln(7)=(3-x)\ln(5)$

18. anonymous

now use calculator to find values of ln7 and ln5

19. anonymous

ln7=1.946 ln5=1.609

20. anonymous

ln(7)=1.94 ln(5)=1.60 (5x-2)*(1.69)=(3-x)1.609

21. anonymous

can you solve this !!

22. anonymous

Yesss. I can take it from there, thank you so much! And the last one, so sorry, is b on the question.

23. anonymous

a typo ! correct is (5x-2)*(1.94)=(3-x)1.609

24. anonymous

ok. u mean this $\Large \sqrt{3x-2}-\sqrt{x}=1$ ??

25. anonymous

Correct :)

26. anonymous

ok taking square of both sides

27. anonymous

Waaiit.

28. anonymous

we have $\Large (\sqrt{3x-2}-\sqrt{x})^2=1^2$ using $\Large (a-b)^2=a^2+b^2-2ab$ we have $\Large ((3x-2)+x-2\sqrt{3x-2}\sqrt{x})=1$

29. anonymous

Sorry, it was 3x-3

30. anonymous

ok

31. anonymous

Itsokay i will pretend the 2 was really 3

32. anonymous

ok

33. anonymous

Sorry :(

34. anonymous

its ok :) i am notorious for typo mistakes ask anyone here :P

35. anonymous

have a look at here http://www.wolframalpha.com/input/?i=solve%28sqrt%283x-3%29-sqrt%28x%29%3D1%29 its quite difficult to type :P

36. anonymous

Wow, thats good enough! Thank you soo much for your help. I learned a lot!

37. anonymous

Also, thank you too @GregTheo

38. anonymous

yw:)

39. anonymous

$\sqrt(3x-2)-\sqrt(x)=1$ add $\sqrt(x)$ in both sides