Question

Solve the following equation:
${{\left( \sqrt{x} \right)}^{{{\log }_{5}}x-1}}=5$

Answer 1

Hint: First simplify it using the laws of exponents. Then put $\log $ and apply the properties of logarithm. After that you will get a quadratic equation to solve it. You will get the answer.
Complete step-by-step answer:
You must have come across the expression ${{3}^{2}}$. Here $3$ is the base and $2$ is the exponent. Exponents are also called Powers or Indices. The exponent of a number tells how many times to use the number in a multiplication. Let us study the laws of the exponent. It is very important to understand how the laws of exponents' laws are formulated.
Product law: According to the product law of exponents when multiplying two numbers that have the same base then we can add the exponents.
${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$
Where a, m, and n all are natural numbers. Here the base should be the same in both quantities.
Quotient Law: According to the quotient law of exponents, we can divide two numbers with the same base by subtracting the exponents. In order to divide two exponents that have the same base, subtract the power in the denominator from the power in the numerator.
$\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$
Power Law: According to the power law of exponents, if a number raises a power to a power, just multiply the exponents.
${{({{a}^{m}})}^{n}}={{a}^{mn}}$
Exponential form refers to a numeric form that involves exponents. One method to write such a number is by identifying that each position is representing a power (exponent) of $10$. Thus, you can initially break it up into different pieces. Exponents are also called Powers or Indices.
In the question we have to find the value of $x$.
We have been given the equation,
${{\left( \sqrt{x} \right)}^{{{\log }_{5}}x-1}}=5$
Now squaring above equation we get,
${{\left( {{(\sqrt{x})}^{2}} \right)}^{{{\log }_{5}}x-1}}={{5}^{2}}$
${{\left( x \right)}^{{{\log }_{5}}x-1}}={{5}^{2}}$
Now using quotient law we get,
$\dfrac{{{\left( x \right)}^{{{\log }_{5}}x}}}{x}={{5}^{2}}$
Now applying $\log $ from both sides we get,
$\log \left( \dfrac{{{\left( x \right)}^{{{\log }_{5}}x}}}{x} \right)=\log {{5}^{2}}$
We know the property of $\log $,
$\begin{align}
  & \log (ab)=\log a+\log b \\
 & \log {{a}^{2}}=2\log a \\
 & \log \left( \dfrac{a}{b} \right)=\log a-\log b \\
 & {{\log }_{a}}b=\dfrac{\log b}{\log a} \\
\end{align}$
So using above properties we get,
\[\begin{align}
  & \log {{(x)}^{{{\log }_{5}}x}}-\log x=\log {{5}^{2}} \\
 & \Rightarrow {{\log }_{5}}x\log (x)-\log (x)=\log {{5}^{2}} \\
 & \Rightarrow \dfrac{\log x}{\log 5}(\log x)-\log x=2\log 5 \\
\end{align}\]
\[\begin{align}
  & \dfrac{{{\left( \log x \right)}^{2}}}{\log 5}-\log x=2\log 5 \\
 & {{\left( \log x \right)}^{2}}-\log x\times \log 5=2{{\left( \log 5 \right)}^{2}} \\
\end{align}\]
\[{{\left( \log x \right)}^{2}}-\log x\times \log 5-2{{\left( \log 5 \right)}^{2}}=0\]
We can see here that there is a quadratic equation formed.
Let $\log x=a$,
So equation becomes,
\[{{a}^{2}}-a\log 5-2{{\left( \log 5 \right)}^{2}}=0\]
We know root of equation $=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$,
Here, we get,
$\begin{align}
  & a=\dfrac{\log 5\pm \sqrt{{{(\log 5)}^{2}}-4(1)(-2{{(\log 5)}^{2}})}}{2(1)} \\
 & a=\dfrac{\log 5\pm \sqrt{{{(\log 5)}^{2}}+8{{(\log 5)}^{2}}}}{2(1)} \\
 & a=\dfrac{\log 5\pm \sqrt{9{{(\log 5)}^{2}}}}{2(1)} \\
 & a=\dfrac{\log 5\pm 3\log 5}{2} \\
\end{align}$
We get,
$a=\dfrac{\log 5+3\log 5}{2}$ or$a=\dfrac{\log 5-3\log 5}{2}$
We get $a$ as,
$a=2\log 5,-\log 5$
Now re substituting we get,
$\log x=\log 25$ and $\log x=\log \left( \dfrac{1}{5} \right)$
So we get the value of $x$ as,
$x=25,\dfrac{1}{5}$
So the values of $x$ are $25$and $\dfrac{1}{5}$.

Note: Read the question carefully. Understand the question and think in the way you can solve it. You must know all the properties of exponents. Also, you must know the properties of logarithm. Do not jumble between the numbers.