Which of the following statements is/are correct?

A. If $2^x = 3^y = 6^{-z}$, then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

B. $(243)^{0.16} \times (9)^{0.1} = 0.3$

C. If $3.105 \times 10^P = 0.00239 + 0.000715$, then P = -3

It is best to first check the (B) statement from the given options.

B. $(243)^{0.16} \times (9)^{0.1} = 0.3$

LHS = $(3)^{5 \times{0.16}} \times (3)^{2 \times {0.1}} = 3^{(0.80)} \times 3^{(0.20)} = 3^1 = 3$

Thus, B is an incorrect statement.

Hence, the only possible option is Option 4.

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A. If $2^x = 3^y = 6^{-z}$, then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

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

Let: $2^{x}=3^{y}=6^{-z}=k \quad \text { (say) }$

Now take logs:

• From $2^{x}=k$, we get $x=\log _{2} k$
  • From $3^{y}=k$, we get $y=\log _{3} k$
  • From $6^{-z}=k$, take log: $-z=\log _{6} k \Rightarrow z=-\log _{6} k$

Now evaluate:

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{\log _{2} k}+\frac{1}{\log _{3} k}+\frac{1}{-\log _{6} k}$

Use change of base: $\frac{1}{\log _{b} k}=\log _{k} b$

So: $=\log _{k} 2+\log _{k} 3-\log _{k} 6=\log _{k}\left(\frac{2 \times 3}{6}\right)=\log _{k}(1)=0$

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C. $3.105 \times 10^{P}=0.00239+0.000715=0.003105$

Thus, $3.105 \times 10^{P}=0.003105 \Rightarrow 10^{P}=\frac{0.003105}{3.105}=10^{-3} \Rightarrow P=-3$