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

- anonymous

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

- katieb

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- anonymous

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

- anonymous

ill use base e

- 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

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

- anonymous

etc ...

- anonymous

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

- 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

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

how do you solve it

- anonymous

it's roughly 68.1257..

- anonymous

how did you get that

- 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

the b is below log

- anonymous

you have to get an approximate also

- anonymous

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

- anonymous

I only come on when I am stuck on problems

- 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

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

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

if you solve this one I will give you a medal

- 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

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

- anonymous

yes

- anonymous

okay - you just need to remember one simple fact:
\[\log_{b}(x) = {\log_{k}(x) \over \log_{k}(b) }

- anonymous

I have got this one wrong 5 times also

- anonymous

\[\log_{b}(x) = {\log_{k}(x) \over \log_{k}(b) } \]

- anonymous

okay what do I plug in where

- anonymous

I got 2.808

- 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

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

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

- anonymous

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

- 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

okay thank you

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