Question

how to write this in single logaritham ln(x/x-5)+ln(x+5/x)-ln(x^2-25)

Answer 1

ln(x/x-5)+ln(x+5/x)-ln(x^2-25) = ln((x-5)/(x+5)) - ln ((x-5)(x+5) = ln(1/(x+5)^2) = ln((x+5)^-2) = -2ln(x+5)

Answer 2

I don't think so. wait

Answer 3

sorry, i think it should be -2ln(x-5), not plus 5

Answer 4

\[
\ln\left(x\over x-5\right)+\ln\left(x+5\over x\right)-\ln(x^2-25)
\]
when logs add, the paranthesis multiply
when logs subtract, they divide..
\[\ln(a)+\ln(b)=\ln(a\times b)\\
\ln(a)-\ln(b)=\ln\left(a\over b\right)\]

Answer 5

they know it. me,too. the problem is a little bit complex. give out the result, please.

Answer 6

i think it is = 2 ln(x-5)

Answer 7

@Hoa try to put your steps here so we can all follow

Answer 8

for the first group, I have \[\ln (x+5) -\ln (x-5)\]

Answer 9

the second one is ln (x+5) +ln(x-5) . cancel out the -ln(x-5) and ln(x-5) you just have ln (x+5) +ln (x+5) =2ln(x+5)

Answer 10

I did the baby step for getting the answer. no need to apply your formula, heheehe

Answer 11

we are putting things together!
\[
\ln\left({x\over x-5}\times{x+5\over x}\times{1\over x^2-25}\right)\\
\qquad=\ln\left(\frac{x+5}{(x-5)(x+5)(x-5)}\right)\\
\qquad=\ln\left(1\over (x-5)^2\right)\\
\qquad=-\ln(x-5)^2\\
\qquad=-2\ln(x-5)
\]