Question

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

Answer 1

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) \end{array}$$ Now if we multiply equation (i) with x, equation (ii) with y and equation(iii) with z We will get its as $$\begin{array}{l} x\log x = kx(y - z)\\\ y\log y = ky(z - x)\\\ z\log z = kz(x - y) \end{array}$$ So now let us add all three of them and we will get it as $$\begin{array}{l} \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 \end{array}$$ Now by using the property of log, $$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.