Question

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

Answer 1

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

Answer 2

ill use base e

Answer 3

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

Answer 4

ln [ 4^(x+4) ] = ln [ 5^(x-6) ]

now use log laws to bring power down in front

Answer 5

etc ...

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

how do you solve it

Answer 10

it's roughly 68.1257..

Answer 11

how did you get that

Answer 12

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

Answer 13

the b is below log

Answer 14

you have to get an approximate also

Answer 15

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

Answer 16

I only come on when I am stuck on problems

Answer 17

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

Answer 18

there's a typo in the second line - it's supposed to be:

(x+4) log 4 = (x-6) log 5

sry, about that...

Answer 19

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

Answer 20

if you solve this one I will give you a medal

Answer 21

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

Answer 22

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

like this?

Answer 23

yes

Answer 24

okay - you just need to remember one simple fact:

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

Answer 25

I have got this one wrong 5 times also

Answer 26

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

Answer 27

okay what do I plug in where

Answer 28

I got 2.808

Answer 29

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...

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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.

Answer 34

okay thank you