Question

If
${a}^{x} = {b}^{y } = {c}^{z}$
and
$abc = 1$
then what is the value of
$\frac{1}{x} + \frac{1}{y } + \frac{1}{z}$
​

Answer 1

Given that,

$\leadsto\ &a^x=b^y=c^z\\ \\ \leadsto\ abc=1$

We have to find,

$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$

Using the identities,

$1.\ \ \log_ab=\dfrac{\log b}{\log a}\\ \\ \\ 2.\ \ \log_m(a)+\log_m(b)+\log_m(c)=\log_m(abc)\\ \\ \\ 3.\ \ \log_a1=0\ \ \ \Longleftarrow\ \ \ a^0=1$

So,

$\textsf{Let}\ \ a^x=b^y=c^z=\mathbf{k}$

$a^x=k\ \ \ \Longrightarrow\ \ \ x=\log_ak\ \ \ \Longrightarrow\ \ \ x=\dfrac{\log k}{\log a}\ \ \ \Longrightarrow\ \ \ \dfrac{1}{x}=\dfrac{\log a}{\log k}\\ \\ \\ b^y=k\ \ \ \Longrightarrow\ \ \ y=\log_bk\ \ \ \Longrightarrow\ \ \ y=\dfrac{\log k}{\log b}\ \ \ \Longrightarrow\ \ \ \dfrac{1}{y}=\dfrac{\log b}{\log k}\\ \\ \\ c^z=k\ \ \ \Longrightarrow\ \ \ z=\log_ck\ \ \ \Longrightarrow\ \ \ z=\dfrac{\log k}{\log c}\ \ \ \Longrightarrow\ \ \ \dfrac{1}{z}=\dfrac{\log c}{\log k}$

Now,

$\begin{aligned}&\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\\ \\ \Longrightarrow\ \ &\frac{\log a}{\log k}+\frac{\log b}{\log k}+\frac{\log c}{\log k}\\ \\ \Longrightarrow\ \ &\frac{\log a+\log b+\log c}{\log k}\\ \\ \Longrightarrow\ \ &\frac{\log (abc)}{\log k}\\ \\ \Longrightarrow\ \ &\frac{\log 1}{\log k}\\ \\ \Longrightarrow\ \ &\frac{0}{\log k}\\ \\ \Longrightarrow\ \ &\Large \textbf{0}\end{aligned}$

Hence,

$\Large\boxed{\begin{minipage}{11.445 cm} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bold{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0} \end{minipage}}$