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[tex]\textbf{Given:}[/tex]
[tex]x=1+log_{a}bc[/tex]
[tex]y=1+log_{b}ca[/tex]
[tex]z=1+log_{c}ab[/tex]
[tex]\textbf{To prove:}[/tex]
[tex]xy+yz+zx=xyz[/tex]
[tex]\textbf{Solution:}[/tex]
[tex]\text{Consider,}[/tex]
[tex]x=1+log_{a}bc[/tex]
[tex]x=log_{a}a+log_{a}bc[/tex]
[tex]\implies\,x=log_{a}abc[/tex]
[tex]\text{similarly}[/tex]
[tex]y=log_{b}abc[/tex]
[tex]z=log_{c}abc[/tex]
[tex]x=log_{a}abc\,\implies\,a^x=abc\,\implies\,a=(abc)^{\frac{1}{x}}[/tex]
[tex]y=log_{b}abc\,\implies\,b^y=abc\,\implies\,b=(abc)^{\frac{1}{y}}[/tex]
[tex]z=log_{c}abc\,\implies\,c^z=abc\,\implies\,c=(abc)^{\frac{1}{z}}[/tex]
[tex]\text{Now, we have}[/tex]
[tex]\bf\,a=(abc)^{\frac{1}{x}}[/tex]
[tex]\bf\,b=(abc)^{\frac{1}{y}}[/tex]
[tex]\bf\,c=(abc)^{\frac{1}{z}}[/tex]
[tex]\text{Multiplying these 3 equations, we get}[/tex]
[tex]abc=(abc)^{\frac{1}{x}}(abc)^{\frac{1}{y}}(abc)^{\frac{1}{z}}[/tex]
[tex]abc=(abc)^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}[/tex]
[tex]abc=(abc)^{\frac{yz+xz+xy}{xyz}}[/tex]
[tex]\text{Equating powers on bothsides, we get}[/tex]
[tex]\dfrac{xy+yz+zx}{xyz}=1[/tex]
[tex]\implies\bf\,xy+yz+zx=xyz[/tex]
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