If we have an expression as \[\dfrac{{\log x}}{{b - c}} = \dfrac{{\log y}}{{c - a}} = \dfrac{{\log z}}{{a - b}}\], show that
i). \[xyz = 1\]
ii). \[{x^a}{y^b}{z^c} = 1\]
iii). \[{x^{b + c}}{y^{c + a}}{z^{a + b}} = 1\]

Hint: Here, in the given question, three different terms are given equal to each other and on the basis of that, we are asked to prove some given equations to be true. First of all, we will take some other constant variable (let’s say \[p\]) equating the three terms given and then find the value of each variable individually. After that we will start with the left hand side of the equation and reach the right hand side of the equation by simplifying it using applicable identities.

Complete step-by-step solution:
Given that \[\dfrac{{\log x}}{{b - c}} = \dfrac{{\log y}}{{c - a}} = \dfrac{{\log z}}{{a - b}}\]
Let \[\dfrac{{\log x}}{{b - c}} = \dfrac{{\log y}}{{c - a}} = \dfrac{{\log z}}{{a - b}} = p\]
\[ \Rightarrow \log x = p\left( {b - c} \right)\]
Assuming that the \[\log \]function has base of \[10\], take exponential function both sides,
\[\therefore x = {10^{p\left( {b - c} \right)}}\]
Similarly,
\[ \log y = p\left( {c - a} \right) \\
   \Rightarrow y = {10^{p\left( {c - a} \right)}} \],
and
\[\log z = p\left( {a - b} \right) \\
   \Rightarrow z = {10^{p\left( {a - b} \right)}} \]
i). To prove: \[xyz = 1\]
We will continue with the left hand side of the equation.
Proof: L.H.S = \[xyz\]
\[ = {10^{p\left( {b - c} \right)}} \times {10^{p\left( {c - a} \right)}} \times {10^{p\left( {a - b} \right)}}\]
Following the multiplication rule which says \[{m^x} \times {m^y} = {m^{x + y}}\], we get,
L.H.S=\[{10^{p\left( {b - c} \right) + p\left( {c - a} \right) + p\left( {a - b} \right)}}\]
Simplifying it, we get,
L.H.S=\[{10^{pb - pc + pc - pa + pa - pb}}\]
=\[{10^0}\]
= \[1\], which is equal to the right hand side of the equation.
Hence, L.H.S = R.H.S
ii). To prove: \[{x^a}{y^b}{z^c} = 1\]
We will continue with the Left hand Side of the equation.
Proof: L.H.S =\[{x^a}{y^b}{z^c}\]
\[ = {\left( {{{10}^{p\left( {b - c} \right)}}} \right)^a} \times {\left( {{{10}^{p\left( {c - a} \right)}}} \right)^b} \times {\left( {{{10}^{p\left( {a - b} \right)}}} \right)^c}\]
Applying the rule that says, \[{\left( {{m^x}} \right)^y} = {m^{xy}}\], we get,
L.H.S = \[{10^{ap\left( {b - c} \right)}} \times {10^{bp\left( {c - a} \right)}} \times {10^{cp\left( {a - b} \right)}}\]
Simplifying it, we get,
L.H.S = \[{10^{abp - apc}} \times {10^{bpc - abp}} \times {10^{acp - bcp}}\]
Following the multiplication rule which says \[{m^x} \times {m^y} = {m^{x + y}}\], we get,
L.H.S = \[{10^{abp - apc + bpc - abp + acp - bcp}}\]
\[ = {10^0}\]
\[ = 1\], which is equal to the right hand side of the equation.
Hence, L.H.S = R.H.S
iii). To prove: \[{x^{b + c}}{y^{c + a}}{z^{a + b}} = 1\]
We will continue with the Left hand Side of the equation.
Proof: L.H.S = \[{x^{b + c}}{y^{c + a}}{z^{a + b}} = 1\]
\[ = {\left( {{{10}^{p\left( {b - c} \right)}}} \right)^{b + c}} \times {\left( {{{10}^{p\left( {c - a} \right)}}} \right)^{c + a}} \times {\left( {{{10}^{p\left( {a - b} \right)}}} \right)^{a + b}}\]
Applying the rule that says, \[{\left( {{m^x}} \right)^y} = {m^{xy}}\], we get,
L.H.S = \[{10^{p\left( {b + c} \right)\left( {b - c} \right)}} \times {10^{p\left( {c + a} \right)\left( {c - a} \right)}} \times {10^{p\left( {a + b} \right)\left( {a - b} \right)}}\]
Simplifying it using the identity \[\left( {x + y} \right)\left( {x - y} \right) = {x^2} - {y^2}\], we get,
L.H.S = \[{10^{p\left( {{b^2} - {c^2}} \right)}} \times {10^{p\left( {{c^2} - {a^2}} \right)}} \times {10^{p\left( {{a^2} - {b^2}} \right)}}\]
Following the multiplication rule which says \[{m^x} \times {m^y} = {m^{x + y}}\], we get,
L.H.S = \[{10^{p\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}\]
\[ = {10^0}\]
\[ = 1\], which is equal to the right hand side of the equation.
Hence, L.H.S = R.H.S

Note: Generally, if the base of \[\log \] function is not given, we assume it as \[10\], or we can say that the base is \[10\] itself if it’s not mentioned. If natural log function is given then the base will be \[e\]. Although, in the given question, it will make no difference what base is there, given any base, the solution will be similar. It is important that we must remember the exponential rules and identities to solve such types of questions.