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anonymous
 5 years ago
I have a logarithmic problem
solve for x approximate
4^(x+4)=5^(x6)
anonymous
 5 years ago
I have a logarithmic problem solve for x approximate 4^(x+4)=5^(x6)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0take logs both sides ( doesnt matter what base, as long as they are the same )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0take log on both side, then you'll get (x+4)log 4 = (x6)log 5, solve it and get the answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln [ 4^(x+4) ] = ln [ 5^(x6) ] now use log laws to bring power down in front

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't come up with the correct answer I have had it wrong 5 times

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have 4 log 5+4log5 / log5log 4 how do I solve it with the calculator they come up with an approximate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's roughly 68.1257..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have to get an approximate also

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't have any clue how to do it no examples

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I only come on when I am stuck on problems

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[4^{x+4} = 5^{x6}\] \[(x+4)\log4 = (x6)\log6 \] \[x \log(4)+4 \log(4) = x \log(5)  6 \log (5)\] \[x \log (4)  x \log (5) = 6 \log (5)  4 \log (4)\] \[x(\log(4)\log(5)) = 6 \log (5)  4 \log (4)\] \[x = {{6 \log (5)  4 \log (4)} \over {\log (4 )  \log (5)}} \approx 68.1257\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there's a typo in the second line  it's supposed to be: (x+4) log 4 = (x6) log 5 sry, about that...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you'll give me medal then here's the solution: 4^(x+4)=5^(x6) => (x+4)log 4 = (x6)log 5 =>(x+4)/(x6) = log5/log 4 => (x+4)/(x6) = 1.16 => (x+4) = 1.16 (x6) =>1.16x x = 4+1.16 (6) => x = 10.96/0.16 => x = 68.5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you solve this one I will give you a medal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\log_{b} (2) = 0.693\] \[\log_{b} (7) = 1.946\] like this?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay  you just need to remember one simple fact: \[\log_{b}(x) = {\log_{k}(x) \over \log_{k}(b) }

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have got this one wrong 5 times also

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\log_{b}(x) = {\log_{k}(x) \over \log_{k}(b) } \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay what do I plug in where

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b 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 base10 logarithm on your calculator and can therefore calculate teh base15 logarithm of 80...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have tried so many different ways to solve it and emailed my professor and he is no help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, one should do his/her assignment himself. just kidding, don't mind!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, 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.
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