Question

Please solve this logarithm question

Answer 1

HEYA !

Given that

$\frac{ \ln(x)}{ b - c} = \frac{ \ln(y)}{ c - a} = \frac{ \ln(z)}{ a - b}$
Now , Let

$\frac{ \ln(x)}{ b - c} = \frac{ \ln(y)}{ c - a} = \frac{ \ln(z)}{ a - b} = k \\ \\ \Rightarrow \: \: \: \frac{ \ln(x)}{ b - c} = k \\ \\ \Rightarrow \: \: \:\ln(x) = k(b - c) \: \: \: \: \: .....(1)$

Similarly , We get

$\ln(y) = k(c- a) \: \: \: \: \: .....(2)$

and

$\ln(z) = k(a- b) \: \: \: \: \: .....(3)$

On adding eq (1) , (2) and (3) , We get

$\ln(x) + \ln(y) + \ln(z) = k(b - c) + k(c - a) + k(a - b) \\ \\ \ln(x yz) = k(b - c + c - a + a - b) \\ \\ \ln(x yz) = k(0) = 0 \\ \\ \ln(x yz) = \ln(1) \: \: \: \: \: \: \: \{ \: \because \ln(1) = 0 \: \} \\ \\ \Rightarrow \: \: \: \boxed{ xyz = 1} \: \: \: \: \: \: \: \: \: \textbf{Ans.}$

Answer 2

Answer:

hai ye sjsjjsssosos madarchod