Question

The product of all the solution of the equation ${{x}^{1+{{\log }_{10}}x}}=100000x$
(a) 10
(b) ${{10}^{5}}$
(c) ${{10}^{-5}}$
(d) 1

Answer 1

Hint: In this question, we have to deal with powers, therefore we know that we will have to concepts of the logarithm. First, we will write the 100000 on the right-hand side as ${{10}^{5}}$. Then we will try to find all the valid values of x. Once we find those values, we will find the product of all those values. That will be the product of the roots.

Complete step-by-step solution:
The equation given to us is ${{x}^{1+{{\log }_{10}}x}}=100000x$.
We will write the 100000 on the right-hand side as ${{10}^{5}}$.
The equation changes as ${{x}^{1+{{\log }_{10}}x}}={{10}^{5}}x$
We also know that ${{x}^{a+b}}={{x}^{a}}.{{x}^{b}}$
Therefore, the left-hand side of the equation changes as $x.{{x}^{{{\log }_{10}}x}}$.
Now, we will take all the terms to the left hand side and take x as common.
$\begin{align}
  & \Rightarrow x.{{x}^{{{\log }_{10}}x}}-{{10}^{5}}x=0 \\
 & \Rightarrow x\left( {{x}^{{{\log }_{10}}x}}-{{10}^{5}} \right)=0 \\
\end{align}$
But we know that $x\ne 0$ as logarithm is not defined for 0 and we have ${{\log }_{10}}x$ as one of the powers.
Therefore, we can safely say that ${{x}^{{{\log }_{10}}x}}-{{10}^{5}}=0$.
We will now try to find all the value of x. For that, we will take the constant term on the right-hand side.
$\Rightarrow {{x}^{{{\log }_{10}}x}}={{10}^{5}}$
Now, if we will take ${{\log }_{x}}$ on both sides.
$\Rightarrow {{\log }_{x}}{{x}^{{{\log }_{10}}x}}={{\log }_{x}}{{10}^{5}}$
It is to be kept in mind that $\log {{x}^{a}}=a\log x$ and another property for logarithm is ${{\log }_{a}}a=1$.
We will use these two properties and simplify our equation.
$\begin{align}
  & \Rightarrow {{\log }_{10}}x{{\log }_{x}}x={{\log }_{x}}{{10}^{5}} \\
 & \Rightarrow {{\log }_{10}}x=5{{\log }_{x}}10 \\
\end{align}$
Now, one other property of logarithm is as follows: ${{\log }_{b}}a=\dfrac{{{\log }_{c}}a}{{{\log }_{c}}b}$, where c is any constant. We will take c as 10 and change ${{\log }_{x}}10$ as $\dfrac{{{\log }_{10}}10}{{{\log }_{10}}x}$.
Therefore, ${{\log }_{10}}x=5\dfrac{{{\log }_{10}}10}{{{\log }_{10}}x}$
We will again use the property ${{\log }_{a}}a=1$ for the numerator on the right hand side.
${{\log }_{10}}x$ gets cross multiplied to the left hand side.
$\begin{align}
  & \Rightarrow {{\log }_{10}}x=5\dfrac{1}{{{\log }_{10}}x} \\
 & \Rightarrow {{\left( {{\log }_{10}}x \right)}^{2}}=5 \\
\end{align}$
We will take square root on both sides.
$\Rightarrow {{\log }_{10}}x=\pm \sqrt{5}$
Therefore, we can say that $x={{10}^{\pm \sqrt{5}}}$.
Therefore, the two roots are $x={{10}^{\sqrt{5}}}$ and $x={{10}^{-\sqrt{5}}}$.
Thus, product of the roots is given as ${{10}^{\sqrt{5}}}\times {{10}^{-\sqrt{5}}}$.
We know that ${{a}^{b}}.{{a}^{c}}={{a}^{b+c}}$.
$\begin{align}
  & \Rightarrow {{10}^{\sqrt{5}-\sqrt{5}}} \\
 & \Rightarrow {{10}^{0}} \\
 & \Rightarrow 1 \\
\end{align}$
Therefore, the product of the roots is 1.
Hence, option (d) is the correct option.

Note: Students should be able to recollect all the properties of the logarithm to solve this question. It is to be noted logarithm is not defined for 0. That is $\log 0$ does not exist. Suppose if ${{\log }_{a}}0=b$, then ${{a}^{b}}$ must be equal to 0 and this is possible only when a = 0, but the base cannot be 0 because 0 raised to power anything is 0. So, there is no definite value of b and thus, $\log 0$ is not defined.