Question

How do I know the restrictions on this log function?
I need to find its solutions which are... 6 and -2 but there is only one due to restrictions, but which one?

Answer 1

\[\log_7(x-1)+\log_7(x-5)\]

Answer 2

what is the domain of \(\log u\) ?

Answer 3

I dont know

Answer 4

For tha base function is more than 0

Answer 5

what does \(\log_by=x\) mean?
it means that \(b^x=y\), which means that if \(b>0\) and \(y<0\) there is no solution
so \(y\) (the thing you are taking the log of, aka the 'argument') must be greater than zero
which of your answers causes one of the arguments to be less than zero?

Answer 6

6

Answer 7

I think I get it now...

Answer 8

let's see\[\log_7(6-1)+\log_7(6-5)=\log_7(5)+\log_7(1)\]none of the arguments are less than zero, so this one is ok

Answer 9

try the other one

Answer 10

it doesnt work

Answer 11

correct\[\log_7(-2-1)+\log_7(-2-5)=\log_7(-3)+\log_7(-3)=undefined\]

Answer 12

correction:\[\log_7(-2-1)+\log_7(-2-5)=\log_7(-3)+\log_7(-7)=undefined\]

Answer 13

thanks

Answer 14

welcome!

Answer 15

@TuringTest what if both solutions were negative

Answer 16

or both solutions positive

are those senrios possible

Answer 17

it doesn't matter if the *solutions* are negative, what matters is that they don't make any of the *arguments* (the parts inside the log function) negative

Answer 18

but are those cases possible

Answer 19

any number of positive or negative solutions is possible, yes

Answer 20

it depends on the equation you start with; if it is a 5th order equation you could get 5 different solutions, all of which may be negative or positive or whatever.

Answer 21

interesting... thanks

Answer 22

sure thing!