If $2^x=3^y=6^{-z}$ then prove that:$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
Solution 1:
$2^x=3^y=6^{-z}=k $ say, then $2= k^{1\over x},3=k^{1\over y},6=k^{-1\over z}$ now can you go on?
then $k^{-1\over z}=6=2\times 3 = k^{1\over x}\times k^{1\over y}=k^{{1\over x}+{1\over y}}$
Solution 2:
$$2^x = 3^y = 6^{-z} = k $$
so$$x = \log_2k$$ $$ y = \log_3k$$ $$z= -\log_6k$$
so $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \log_k2 + log_k3 -\log_k6$$ $$=\log_k{\frac{2\times3}{6}}$$ $$=0$$