anonymous
  • anonymous
I have a logarithmic problem solve for x approximate 4^(x+4)=5^(x-6)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
take logs both sides ( doesnt matter what base, as long as they are the same )
anonymous
  • anonymous
ill use base e
anonymous
  • anonymous
take log on both side, then you'll get (x+4)log 4 = (x-6)log 5, solve it and get the answer.

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More answers

anonymous
  • anonymous
ln [ 4^(x+4) ] = ln [ 5^(x-6) ] now use log laws to bring power down in front
anonymous
  • anonymous
etc ...
anonymous
  • anonymous
I don't come up with the correct answer I have had it wrong 5 times
anonymous
  • anonymous
I have 4 log 5+4log5 / log5-log 4 how do I solve it with the calculator they come up with an approximate
anonymous
  • 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.
anonymous
  • anonymous
how do you solve it
anonymous
  • anonymous
it's roughly 68.1257..
anonymous
  • anonymous
how did you get that
anonymous
  • 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
anonymous
  • anonymous
the b is below log
anonymous
  • anonymous
you have to get an approximate also
anonymous
  • anonymous
don't have any clue how to do it no examples
anonymous
  • anonymous
I only come on when I am stuck on problems
anonymous
  • anonymous
\[4^{x+4} = 5^{x-6}\] \[(x+4)\log4 = (x-6)\log6 \] \[x \log(4)+4 \log(4) = x \log(5) - 6 \log (5)\] \[x \log (4) - x \log (5) = -6 \log (5) - 4 \log (4)\] \[x(\log(4)-\log(5)) = -6 \log (5) - 4 \log (4)\] \[x = {{-6 \log (5) - 4 \log (4)} \over {\log (4 ) - \log (5)}} \approx 68.1257\]
anonymous
  • 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...
anonymous
  • 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
anonymous
  • anonymous
if you solve this one I will give you a medal
anonymous
  • 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
anonymous
  • anonymous
\[\log_{b} (2) = 0.693\] \[\log_{b} (7) = 1.946\] like this?
anonymous
  • anonymous
yes
anonymous
  • anonymous
okay - you just need to remember one simple fact: \[\log_{b}(x) = {\log_{k}(x) \over \log_{k}(b) }
anonymous
  • anonymous
I have got this one wrong 5 times also
anonymous
  • anonymous
\[\log_{b}(x) = {\log_{k}(x) \over \log_{k}(b) } \]
anonymous
  • anonymous
okay what do I plug in where
anonymous
  • anonymous
I got 2.808
anonymous
  • 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...
anonymous
  • 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
anonymous
  • anonymous
I have tried so many different ways to solve it and emailed my professor and he is no help
anonymous
  • anonymous
lol, one should do his/her assignment himself. just kidding, don't mind!!
anonymous
  • 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.
anonymous
  • anonymous
okay thank you

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