Jaweria 2 years ago Help me Please. e^x-1-5=5.

1. Jaweria

2. yakeyglee

Solve for $$e^{x-1}$$ and take the natural logarithm of both sides.

3. Jaweria

I started this problem but its giving me a wrong answer can u please explain this to me?

4. yakeyglee

Well, tell me, what do you get when you solve for $$e^{x-1}$$?

5. Jaweria

can anyone help me here please?

6. stamp

$e^x-1-5=5.$$e^x-6=5$$e^x=11$$ln(e^x)=ln(11)$$xln(e)=ln(11)$$x=ln(11)$

7. Jaweria

Sorry but the answer is wrong answer suppose to be 3.3026

8. stamp

@Jaweria Then you wrote it down wrong.

9. Jaweria

ok let me fix it

10. Jaweria

$e ^{x-1}-5=5$

11. stamp

$e^{x-1}-5=5$Isolate e term.$e^{x-1}=10$Natural log of both sides.$lne^{x-1}=ln(10)$Properties of logatrithms.$(x-1)lne=ln(10)$lne = 1.$x-1=ln(10)$Solve for x.$x=ln(10)+1$

12. Jaweria

Sorry but I have one more question if you dont mind ?

13. Jaweria

$9^{n+10}+3=81$

14. stamp

$9^{n+10}+3=81$$9^{n+10}=81-3$$ln9^{n+10}=ln78$$n+10=ln78/ln9$$n=ln78/ln9-10$

15. Jaweria

Thanks :)

16. stamp

@Jaweria Properties of Logarithms http://www.andrews.edu/~calkins/math/webtexts/numb17.htm

17. Jaweria

Thanks for this :)

18. Jaweria

I m having trouble with this question. $-2 \log_{5}7x=2$

19. stamp

$-2log_57x=2$$log_57x=-1$$log_ax=y$$a^y=x$$5^{-1}=7x$$\frac{1}{5}=7x$$x=1/35$

20. Jaweria

Thanks

21. Jaweria

I m working on this problem so far I got this but my answer is not matching with my professor's. |dw:1361342182376:dw|

22. Jaweria

23. whpalmer4

Is that $5*6^{3m} = 20$?

24. Jaweria

yes

25. whpalmer4

I'll assume it is, if it isn't, maybe the example will help you. $5*6^{3m} = 20$Divide both sides by 5$6^{3m}=4$Take natural log of both sides$3m \ln 6 = \ln 4$Divide both sides by $$3 \ln 6$$$m = \frac{\ln 4}{3 \ln 6}$

26. Jaweria

thanks