Question 33

If $$2^{x}=3^{y}=6^{-z}$$, then $$\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$$ is equal to

$$2^{x}=3^{y}=6^{-z}$$ = K

X = $$\log_2k$$
Y = $$\log_3k$$
Z = -($$\log_6k$$)

Hence, $$\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$$ = $$\log_k2$$ + $$\log_k3$$ - $$\log_k6$$
                                          = $$\log_k6$$ - $$\log_k6$$
                                          = 0.

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