Question

solve \[\log_{7}(x+1)+\log_{7}(x-5)=1 \]

Answer 1

use base change formula

Answer 2

\[\log _{7}(x+5)=\frac{ \log (x+5) }{ \log 7 }\]

Answer 3

after that use\[\log (a*b)= \log a + \log b\]

Answer 4

@alexeis_nicole can you do it?

Answer 5

its \[\log_{7}(x-5) .. \] not plus :S

Answer 6

sorry, somewhat it will be like \[\log (\frac{ x+1 }{ 7 }*\frac{ x-5 }{ 7 })\]

Answer 7

can you do it now? take the log on the other side it will be e^1=e=2.732 and you will get a quadratic equation, hope you can do it now

Answer 8

i got to log[(x^2-4x-5)/7]=1

Answer 9

@ghazi

Answer 10

now, \[\frac{ (x+1)(x-5) }{ 49 }=e^1=e=2.73\] then multiply both sides by 49

Answer 11

uhh im kinda lost now.. /:the way my book shows it is different from how you are @ghazi

Answer 12

well you just need to use properties of logs that is it or may be you can replace e by 10 if the base of log is 10

Answer 13

\[\frac{ (x+1)(x-5) }{ 49 }=10\] then \[(x+1)(x-5)=490\]

Answer 14

\[\log_{7}((x+1)(x-5))=1\]

(x+1)(x-5)=7^1
x^2-4x-5=7
x^2-4x-12=0
(x-6)(x+2)
x=6 x=-2