## anonymous one year ago Solve for x. logx+log(x-15)=2

1. hartnn

hello :) what did you try?? tried combining those 2 logs?

2. anonymous

I know what you have to do, however, I am not sure how to do it :(

3. anonymous

I don't know how to combine the logs

4. hartnn

$$\log A + \log B = \log AB$$ so how about log x + log (x-15) ?

5. anonymous

is it 20?

6. hartnn

you're asking if x = 20 ? how did you do all the steps so fast! :O

7. anonymous

I've been working on this problem for a really long time and I wasn't sure if I was doing the steps right so I posted on here. I think the answer is 20...I don't know what else it could be :(

8. hartnn

oh great! you've tried some steps... can you post them here, we'll correct then if required.

9. anonymous

Oh boy... I don't know where to begin! I'll write the most recent steps I've tried...one moment

10. anonymous

I used the formula $a=\log _{b}(b ^{a})$

11. anonymous

$2= \log _{10}(10^2)$

12. hartnn

for the right side? so, $$2 = \log 10^2$$

13. hartnn

but for the left side we will need the formula I posted :)

14. hartnn

$$\log x + \log(x-15) = \log x\times(x-15) \\= \log 10^2$$

15. anonymous

$\log _{10}(x)+\log _{10}(x-15)=\log _{10}(10^2)$

16. hartnn

good

17. anonymous

$\log _{10}(x(x-15))=\log _{10}(10^2)$

18. hartnn

you get how $$\Large \log x + \log (x-15) = \log [x\times (x-15)]$$ basically just a log property

19. anonymous

and the answeres are x=20 and x=-5 but we cant have a negative log so it must be 20

20. hartnn

if log A = log B, then A = B what that says is that you can just cancel out log on both sides! so we have x(x-15) = 10^2 got this?

21. anonymous

yes I just skipped the last steps at the end because they were really small and by then I already figured out my answers :)

22. hartnn

oh you solved it completely. excellent! x =20 is correct :)

23. anonymous

Thanks @hartnn , I appreciate your time :)

24. hartnn

welcome ^_^ happy to help :)