Question

If a = logx (yz), b = logy (zx), c = logz (xy) then prove that, 1/(1+a)+1/(1+b)+1/(1+c) =1​

Answer 1

answer to the question is

Answer 2

The proof is given below

Given

$a=\log_x(yz)$          .................. (1)

$b=\log_y(zx)$          .................. (2)

$c=\log_z(xy)$          .................. (3)

From eq (1)

$a+1=\log_x(yz)+1$

$\implies a+1=\log_x(yz)+\log_xx$

$\implies a+1=\log_x(xyz)$

$\implies \frac{1}{a+1}=\frac{1}{\log_xxyz}$

Similarly, from eq (2) and (3)

$\implies \frac{1}{b+1}=\frac{1}{\log_yxyz}$

$\implies \frac{1}{c+1}=\frac{1}{\log_zxyz}$

Thus,

$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$

$=\frac{1}{\log_xxyz}+\frac{1}{\log_yxyz}+\frac{1}{\log_zxyz}$

$=\frac{1}{\log xyz/\log x}+\frac{1}{\log xyz/\log y}+\frac{1}{\log xyz/\log z}$

$=\frac{\log x}{\log xyz}+\frac{\log y}{\log xyz}+\frac{\log z}{\log xyz}$

$=\frac{\log x+\log y+\log z}{\log xyz}$

$=\frac{\log xyz}{\log xyz}$

$=1$                                    (Proved)

Hope this answer is helpful

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