## anonymous 4 years ago solve the equation ln(X-3)=ln(X+2)+3 how do you cancel ln. I know when you have log you can do this= log base 5 (2X+4)=3

1. anonymous

Just a start and i'm not definitive on this but can you do ln( (x-3)/(x+2) ) = 3 as subtracting logs equates to dividing them as one 'ln'.

2. TuringTest

^that is correct then raise e to the power of both sides...

3. anonymous

can you help on this problem, Ln(x-3)=Ln(x+2)+3 please! so i can see it? how you do it

4. TuringTest

As is stated above remember the log rule$\log a-\log b=\log(\frac ab)$so here we can do the following:$\ln(x-3)=\ln(x+2)+3$$\ln(\frac{x-3}{x+2})=3$now raise e to the power of both sides and isolate x:$\frac{x-3}{x+2}=e^3$$x-3=e^3x+2e^3$$x-e^3x=3+2e^3$$x(1-e^3)=3+2e^3$$x=\frac{3+2e^3}{1-e^3}$please pay try to follow how I did this so you can do the problems yourself in the future.

5. TuringTest

6. anonymous

awesome!! thx you yes, i will try myself so i can be ready for my test, right practice take perfection.thank you so much!

7. TuringTest

welcome!

8. anonymous

LnX+Ln(X+2)=2 this is what i did Ln(X+0)/(X+2)=2 (X+0)/(X+2)=e^2 x+=e^2X+2e^2 x+e^2-x=-0+2e^2 x=2e^2/x+e^2= the calculator says error cause i can not divide by o, off course, but my professor book answer is x=1.896 or x=-3.896 what iam doing wrong?

9. TuringTest

you have a slightly different rule applying with addition$\log a+\log b=\log(ab)$so here you have$\ln x+\ln(x+2)=\ln(x^2+2x)$try to take it from there

10. TuringTest

...and you never need to add a zero like that. No point in it.

11. TuringTest

Basic log rules:$\log a+\log b=\log(ab)$$\log a-\log b=\log(\frac ab)$$\log(x^a)=a\log x$$\log_b x=\frac{\log_a x}{\log_a b}$these are your best friends for these kind of problems.

12. anonymous

so, Ln x cancel and just end up with(x+2)/(x^2+2x)?

13. anonymous

oh.. did you multiple (x).(x+2) and end up x^2+2x-2=0 and from then do the quadratic formula?

14. TuringTest

$\ln x+\ln(x+2)=2$$\ln(x^2+2x)=2$$x^2+2x=e^2$$x^2+2x-e^2=0$now you can use the quadratic formula.

15. anonymous

16. anonymous

Take ln of both sides as ln undoes e^x.. so: ln(e^x) = ln(35) x = ln(35)

17. anonymous

oh i was doing e^35 iam a little confused i though the ln cancel in both sides

18. anonymous

how would you do this problem 2^3x=16 i know base is 2

19. anonymous

I'm not 100% sure but if you take log(base 2) on both sides you get 3x = 4 hence x = 4/3 is that correct?

20. anonymous

no, is 2 to the power of 3x=16

21. anonymous

2^3x=16

22. anonymous

yeah, that's what I did - solve that for x which gives you 4/3. type log base 2 (16) into your calculator and you'll get 4. hence 3x = 4

23. anonymous

ah.. thx! what the formula for that problem?

24. anonymous

None - just knowing the rules of logarithms.

25. anonymous

i started this problem original problem is 5e^4x=500 so what i did is the following:5/5 e^4x=500/5 e^4x=e^100 4x=4.60517-4 -4x x=.60517?

26. anonymous

Chances are that's incorrect as answers rarely come out like that. I think this is correct: e^4x = 500/5 e^4x = 100 ln(e^4x) = ln(100) 4x = ln(100) x = ln(25) <-- always leave your answer in terms on ln as it is more precise than a decimal.

27. anonymous

I've made a mistake again.. I'm getting tired: x = ln(100) / 4

28. anonymous

I understand you been tired, I been home work since 11a.m i shall also take a break. thank you so much for you help i really

29. anonymous

No worries.