Alg3.3.11

If $a^x=b^y=c^z=d^w\,$ and $ab=cd\,$ .Show that $\frac{1}{x}+\frac{1}{y}=\frac{1}{w}+\frac{1}{z}\,$

Given $a^x=b^y=c^z=d^w\,$ and $ab=cd\,$ .

From the given first equation,let

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

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

Substituting the values in the given condition $ab=cd\,$ we get

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

$k^{\frac{1}{x}+\frac{1}{y}}=k^{\frac{1}{z}+\frac{1}{w}}\,$

Since bases on both sides are equal

$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}+\frac{1}{w}\,$