Question

Solution set of inequality ${5^{(x + 2)}} > {\left( {\dfrac{1}{{25}}} \right)^{\dfrac{1}{x}}}$
(A) $\left( { - 2,0} \right)$
(B) $\left( { - 2,2} \right)$
(C) $\left( { - 5,5} \right)$
(D) $\left( {0,\infty } \right)$

Answer 1

Hint: We solve this question by first writing 25 as square of 5 and then applying the property, when an exponential equation has the same base on each side then their exponents must also be equal, that is, if ${a^n} > {a^m}$ then, $n > m$. Then, solving the equation to find the solution set of given inequality. We must have knowledge about the exponential and power rules in order to solve the inequality.

Complete step-by-step answer:
Given inequality is ${5^{(x + 2)}} > {\left( {\dfrac{1}{{25}}} \right)^{\dfrac{1}{x}}}$.
First, write 25 as square of 5
${5^{(x + 2)}} > {\left( {\dfrac{1}{{{5^2}}}} \right)^{\dfrac{1}{x}}}$
Now, we take ${5^2}$in numerator which becomes ${5^{ - 2}}$ due to property which states that $\dfrac{1}{{{a^n}}} = {a^{ - n}}$
So, the inequality changes to
${5^{\left( {x + 2} \right)}} = {\left( {{5^{ - 2}}} \right)^{\dfrac{1}{x}}}$
Now multiplying $ - 2$with $\dfrac{1}{x}$ as ${\left( {{a^s}} \right)^r} = {a^{sr}}$
The inequality changes to,
${5^{\left( {x + 2} \right)}} = {\left( 5 \right)^{\dfrac{{ - 2}}{x}}}$
Now, applying the same base property, that, when an exponential equation has the same base on each side then their exponents must also be equal, that is, if ${a^n} > {a^m}$ then, $n > m$.
We get,
$x + 2 > \dfrac{{ - 2}}{x}$
Taking $\dfrac{{ - 2}}{x}$to left hand side, we get,
$x + 2 - \dfrac{2}{x} > 0$
$\dfrac{{{x^2} + 2x - 2}}{x} > 0$
First solving for the discriminant of the quadratic expression and finding the nature of solutions.
On comparing given equation with the general equation $a{x^2} + bx + c = 0$, we get,
\[a = 1\]
$b = 2$
$c = - 2$
Now, we know that the expression for discriminant is $D = {b^2} - 4ac$. So, we have,
$ \Rightarrow D = {2^2} - 4\left( 1 \right)\left( 2 \right)$
$ \Rightarrow D = 4 - 8$
$ \Rightarrow D = - 4$
So, we get the value of discriminant is negative. So, this means that there are no real roots and that it does not touch the x axis when plotted on the Cartesian plane. Also, value of a is positive as \[a = 1\]. So, the parabola represented by the quadratic expression ${x^2} + 2x - 2$ is facing upwards.
Hence, we can say that the expression is positive for every value of x.
Now, coming back to the inequality $\dfrac{{{x^2} + 2x - 2}}{x} > 0$
We know that, ${x^2} + 2x - 2 > 0$ for every value of x , which implies that,
$x > 0$
This means that,
Solution set of inequality ${5^{(x + 2)}} > {\left( {\dfrac{1}{{25}}} \right)^{\dfrac{1}{x}}}$ is $\left( {0,\infty } \right)$
Hence, option (d) is the correct answer.
So, the correct answer is “Option d”.

Note: This question deals with the knowledge of the same basic exponential property, that, , when an exponential equation has the same base on each side then their exponents must also be equal, that is, if ${a^n} > {a^m}$ then, $n > m$ for a greater than one. And also keep in mind the general form of quadratic equation is $a{x^2} + bx + c = 0$. Take care while doing the calculations.