Question

solve this problem!! ​

Answer 1

$\textbf{Given:}$

$x=1+log_{a}bc$

$y=1+log_{b}ca$

$z=1+log_{c}ab$

$\textbf{To prove:}$

$xy+yz+zx=xyz$

$\textbf{Solution:}$

$\text{Consider,}$

$x=1+log_{a}bc$

$x=log_{a}a+log_{a}bc$

$\implies\,x=log_{a}abc$

$\text{similarly}$

$y=log_{b}abc$

$z=log_{c}abc$

$x=log_{a}abc\,\implies\,a^x=abc\,\implies\,a=(abc)^{\frac{1}{x}}$

$y=log_{b}abc\,\implies\,b^y=abc\,\implies\,b=(abc)^{\frac{1}{y}}$

$z=log_{c}abc\,\implies\,c^z=abc\,\implies\,c=(abc)^{\frac{1}{z}}$

$\text{Now, we have}$

$\bf\,a=(abc)^{\frac{1}{x}}$

$\bf\,b=(abc)^{\frac{1}{y}}$

$\bf\,c=(abc)^{\frac{1}{z}}$

$\text{Multiplying these 3 equations, we get}$

$abc=(abc)^{\frac{1}{x}}(abc)^{\frac{1}{y}}(abc)^{\frac{1}{z}}$

$abc=(abc)^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

$abc=(abc)^{\frac{yz+xz+xy}{xyz}}$

$\text{Equating powers on bothsides, we get}$

$\dfrac{xy+yz+zx}{xyz}=1$

$\implies\bf\,xy+yz+zx=xyz$

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