Question

solve a log

Answer 1

\[\log_{a}x+\log_{a}x-2=\log_{a}x+4 \]\
\[\log_{a}x(x-2)=\log_{a}x+4\]
that is my work so far I do not know how to convert to exponential

Answer 2

I am wondering if I divide by \[\log_{a}x+4\] and then have it equal to zero which means my final answer is just 1

Answer 3

assuming the right hand side is \(\log_a(x+4)\) then your last job is to solve
\[x(x-2)=x+4\]

Answer 4

so the logs just drop out

Answer 5

log is a "one to one function" so if
\[\log(A)=\log(B)\] then \(A=B\)

Answer 6

ok I guess but I dont really understand

Answer 7

ok suppose instead of log, you had
\[f(x)=2x+4\]

Answer 8

if you were told that
\[2a+4=2b+4\] then you would know that \(a=b\) right?

Answer 9

it is the same with the log
if \(\log(x)=\log(y)\) then you know \(x=y\)

Answer 10

in other words, as in your example, the only way for
\(\log(x^2-2x)\) to be equal to \(\log(x+4)\) is if \(x^2-2x=x+4\)

Answer 11

oh I see now so if i understand correctly it is because we know they must be equal that the fact that it is a log of each does not matter because no matter what they yield the same result

Answer 12

yes, exactly

Answer 13

Thank you I appreciate your help

Answer 14

if i know \(\log(x)=\log(5)\) then i know that \(x=5\) i cannot be anything else

Answer 15

yw