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If we have a logarithmic expression \[\log x:\log y:\log z = (y - z):(z - x):(x - y)\] then which of the following expressions is true?
A). \[{x^x}.{y^y}.{z^z} = 1\]
B). \[{x^y}.{y^z}.{z^x} = 1\]
C). \[\sqrt[x]{x}\sqrt[y]{y}\sqrt[z]{z} = 1\]
D). None of these

Last updated date: 16th Jun 2024
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Hint: This is a typical logarithmic property question. We will need to use the property \[x\log x = \log \left( {{x^x}} \right)\] to get the final answer. Also we will need some properties of ratio to get started.

Complete step-by-step answer:
It is given that \[\log x:\log y:\log z = (y - z):(z - x):(x - y)\]
Then we can let that,
\[\dfrac{{\log x}}{{y - z}} = \dfrac{{\log y}}{{z - x}} = \dfrac{{\log z}}{{y - z}} = k\]
So from here we can get
\log x = k(y - z)...................................(i)\\
\log y = k(z - x)................................(ii)\\
\log z = k(x - y).................................(iii)
Now if we multiply equation (i) with x, equation (ii) with y and equation(iii) with z
We will get its as
x\log x = kx(y - z)\\
y\log y = ky(z - x)\\
z\log z = kz(x - y)
So now let us add all three of them and we will get it as
\therefore x\log x + y\log y + z\log z = kx(y - z) + ky(z - x) + kz(x - y)\\
 \Rightarrow x\log x + y\log y + z\log z = k\left( {xy - xz + yz - xy + yz + xz - yz} \right)\\
 \Rightarrow x\log x + y\log y + z\log z = k \times 0\\
 \Rightarrow x\log x + y\log y + z\log z = 0
Now by using the property of log, \[x\log x = \log \left( {{x^x}} \right)\] we will get the whole thing as
\[ \Rightarrow \log \left( {{x^x}} \right) + \log \left( {{y^y}} \right) + \log \left( {{z^z}} \right) = 0\]
Now again we know that \[\log x + \log y + \log z = \log (xyz)\]
So by using this property of log we are getting
\[ \Rightarrow \log \left( {{x^x}.{y^y}.{z^z}} \right) = 0\]
Again if we take the log to the other side we will get it as
\[ \Rightarrow \left( {{x^x}.{y^y}.{z^z}} \right) = 1\]
Which is the correct answer and which means that option A is the correct option here.

Note: \[\log x = 0\] is \[x = 1\] because we know that logarithm in mathematics is treated as log base e so here \[{\log _e}x = 0\] is basically \[x = {e^0} = 1\] . In other words we can also say that the antilog of 0 is 1.