Question

Logarithms

Answer 1

\[\frac{ logb }{ b-c }=\frac{ logb }{ c-a }=\frac{ logc }{ a-b }\]
Then
\[a ^{b+c} . b ^{c+a} . c^{a+b}\]
equals

Answer 2

Confirm: Is the numerator of the first term log b or log a?

Answer 3

log a sorry

Answer 4

is this log base e?

Answer 5

kainui i did the derivative what now?

Answer 6

I'm not sure whether it's right or wrong. I just give out my opinion. :)
\[\frac{ loga }{ b-c }=\frac{ logb }{ c-a }\\log_b^a=\dfrac{b-c}{c-a}\rightarrow b^\dfrac{b-c}{c-a} =a \\a^{b+c}=b^{\dfrac{b^2-c^2}{c-a} }\]
by that way, I can calculate \(b^{c+a}\) and \(c^{a+b}\) then times them together.