## anonymous 5 years ago I have a logarithmic problem solve for x approximate 4^(x+4)=5^(x-6)

1. anonymous

take logs both sides ( doesnt matter what base, as long as they are the same )

2. anonymous

ill use base e

3. anonymous

take log on both side, then you'll get (x+4)log 4 = (x-6)log 5, solve it and get the answer.

4. anonymous

ln [ 4^(x+4) ] = ln [ 5^(x-6) ] now use log laws to bring power down in front

5. anonymous

etc ...

6. anonymous

I don't come up with the correct answer I have had it wrong 5 times

7. anonymous

I have 4 log 5+4log5 / log5-log 4 how do I solve it with the calculator they come up with an approximate

8. anonymous

I don't have my calculator right now with me, nor my internet connection is working fine and this site is damn slow or else i would have solved that for you.

9. anonymous

how do you solve it

10. anonymous

it's roughly 68.1257..

11. anonymous

how did you get that

12. anonymous

I have another one you might be able to help me with use log b 2=0.693 and /or lo b 7 =1.946 to find log b 14

13. anonymous

the b is below log

14. anonymous

you have to get an approximate also

15. anonymous

don't have any clue how to do it no examples

16. anonymous

I only come on when I am stuck on problems

17. anonymous

$4^{x+4} = 5^{x-6}$ $(x+4)\log4 = (x-6)\log6$ $x \log(4)+4 \log(4) = x \log(5) - 6 \log (5)$ $x \log (4) - x \log (5) = -6 \log (5) - 4 \log (4)$ $x(\log(4)-\log(5)) = -6 \log (5) - 4 \log (4)$ $x = {{-6 \log (5) - 4 \log (4)} \over {\log (4 ) - \log (5)}} \approx 68.1257$

18. anonymous

there's a typo in the second line - it's supposed to be: (x+4) log 4 = (x-6) log 5 sry, about that...

19. anonymous

if you'll give me medal then here's the solution: 4^(x+4)=5^(x-6) => (x+4)log 4 = (x-6)log 5 =>(x+4)/(x-6) = log5/log 4 => (x+4)/(x-6) = 1.16 => (x+4) = 1.16 (x-6) =>1.16x -x = 4+1.16 (6) => x = 10.96/0.16 => x = 68.5

20. anonymous

if you solve this one I will give you a medal

21. anonymous

use log b 2=0.693 and /or log b7=1.946 to find log b 14 log b 14 = approximate the b is below log

22. anonymous

$\log_{b} (2) = 0.693$ $\log_{b} (7) = 1.946$ like this?

23. anonymous

yes

24. anonymous

okay - you just need to remember one simple fact: $\log_{b}(x) = {\log_{k}(x) \over \log_{k}(b) } 25. anonymous I have got this one wrong 5 times also 26. anonymous \[\log_{b}(x) = {\log_{k}(x) \over \log_{k}(b) }$

27. anonymous

okay what do I plug in where

28. anonymous

I got 2.808

29. anonymous

b is your unknown base and k is just some random base you can choose... for example $\log_{15} (80) = { \log_{10}(80) \over \log_{10}(15) }$ so you got the base-10 logarithm on your calculator and can therefore calculate teh base-15 logarithm of 80...

30. anonymous

I just don't get it can you help me this please solve I just want to get this BS done it is my last problem

31. anonymous

I have tried so many different ways to solve it and emailed my professor and he is no help

32. anonymous

lol, one should do his/her assignment himself. just kidding, don't mind!!

33. anonymous

okay, okay - BUT, you rly should try to understand it, cause it's quite important. $\log_{b}(2) = {\ln (2) \over \ln(b)} = 0.693$ $\ln(b) = {\ln(2) \over 0.693}$ $b = e ^ {\ln(2) \over 0.693} \approx e ^ 1 \approx 2.7183$ the last line is due to the fact, that ln(2) is about 0.6931... now you should try the same with the other fact you got and check the result.

34. anonymous

okay thank you