Question

Solve the following inequality:
$-\log _{4}^{2}x+13{{\log }_{4}}x-36 > 0$

Answer 1

Hint: First of all, let us assume that ${{\log }_{4}}x=t$ then we are going to substitute in the given inequality and we get $-{{t}^{2}}+13t-36 > 0$.Now, multiplying -1 on both the sides then the sign of the inequality will change and we have ${{t}^{2}}-13t+36 < 0$. We are going to solve the quadratic expression in “t” by factorization method and then find the range of the value of t. After that put ${{\log }_{4}}x=t$ and find the range of x.

Complete step-by-step solution -
The inequality given in the question is:
$-\log _{4}^{2}x+13{{\log }_{4}}x-36 > 0$
Let us assume the ${{\log }_{4}}x=t$ and plugging this value of ${{\log }_{4}}x$ as “t” in the above inequality we get,
$-{{t}^{2}}+13t-36 > 0$
Multiplying the above inequality by -1, the sign of inequality will change and we get,
${{t}^{2}}-13t+36 < 0$
The above inequality is quadratic in “t” so we are going to factorize the quadratic expression as follows:
$\begin{align}
  & {{t}^{2}}-9t-4t+36 < 0 \\
 & \Rightarrow t\left( t-9 \right)-4\left( t-9 \right) < 0 \\
 & \Rightarrow \left( t-4 \right)\left( t-9 \right) < 0 \\
\end{align}$
Now, we have to find the range of “t” in such a manner that multiplication of t – 4 and t – 9 must be less than 0 so if we put the value of the “t” as less than 4 in $\left( t-4 \right)\left( t-9 \right)$ then the sign of this multiplicative expression is positive or greater than 0 so we cannot take values which are less than 4. Now, if we put the value of “t” between 4 and 9 then multiplicative expression $\left( t-4 \right)\left( t-9 \right)$ will have negative value means less than 0 which is acceptable because in the above, we have the multiplication of the expression $\left( t-4 \right)\left( t-9 \right)$ is less than 0. And if we put the value greater than 9 then the multiplicative expression gives a value greater than 0 which is not acceptable.
So, from the above discussion we have found that the acceptable range of “t” is between 4 to 9 and excluding 4 and 9 because at 4 and 9 the multiplicative expression becomes 0.
$t\in \left( 4,9 \right)$
As we have assumed above that ${{\log }_{4}}x=t$ so equating ${{\log }_{4}}x$ to 4 we get,
${{\log }_{4}}x=4$
Taking antilog on both the sides we get,
$x={{4}^{4}}=256$
Equating ${{\log }_{4}}x$ to 9 we get,
${{\log }_{4}}x=9$
Taking antilog on both the sides of the above equation we get,
$\begin{align}
  & x={{9}^{4}} \\
 & \Rightarrow x=6561 \\
\end{align}$
From the above, the range of x is equal to:
$x\in \left( 256,6561 \right)$
The brackets above show that 256 and 6561 value is not included in the range of x.
Hence, the range of x is equal to $x\in \left( 256,6561 \right)$.

Note: You might think how we have factorize the quadratic expression in the above solution.
The quadratic expression in the above solution that we have got is:
${{t}^{2}}-13t+36 < 0$
We are only going to show how we have factorized the quadratic expression in “t”.
${{t}^{2}}-13t+36$
First of all write the factors of 36 which are:
$36=3\times 3\times 2\times 2\times 1$
Now, add or subtract the factors of 36 in such a way that it will give the number 13 so if we add the factors of 36 i.e. 9 and 4 we will get 13.
We are going to write 13 as 9 + 4 in the above quadratic expression we get,
$\begin{align}
  & {{t}^{2}}-13t+36 \\
 & ={{t}^{2}}-\left( 9+4 \right)t+36 \\
 & ={{t}^{2}}-9t-4t+36 \\
\end{align}$
Taking “t” as common from the first two terms of the above expression and taking -4 as common from the last two terms of the above expression we get,
$t\left( t-9 \right)-4\left( t-9 \right)$
As you can see from the above expression that $\left( t-9 \right)$ is common so we can take the $\left( t-9 \right)$out from this expression and write the remaining terms.
$\left( t-9 \right)\left( t-4 \right)$
This is how we have done the factorization of the quadratic expression.