Alg9.4.44

If $a^x=b^y=c^z=d^w\,$ Show that $\log_a (bcd)=x\cdot(\frac{1}{y}+\frac{1}{z}+\frac{1}{w})\,$

Let

$a^x=b^y=c^z=d^w=k\,$

$a=k^{\frac{1}{x}},b=k^{\frac{1}{y}},c=k^{\frac{1}{z}},d=k^{\frac{1}{w}}\,$

Substituting the values in the left expression,we get

$\log_a(bcd)=\log_{k^{\frac{1}{x}}}(k^{\frac{1}{y}}\cdot k^{\frac{1}{z}}\cdot k^{\frac{1}{w}})\,$

$\log_a(bcd)=\log_{k^{\frac{1}{x}}}(k^{\frac{1}{y}+\frac{1}{z}+\frac{1}{w}})\,$

Hence

$x\cdot(\frac{1}{y}+\frac{1}{z}+\frac{1}{w})\,$