Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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
Total views: 412.2k
Views today: 10.12k
Answer
Verified
412.2k+ views
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
$\begin{array}{l} \log x = k(y - z)...................................(i)\\ \log y = k(z - x)................................(ii)\\ \log z = k(x - y).................................(iii) \end{array}$
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.