Question

Solve- ${(\dfrac{2}{5})^{\dfrac{{6 - 5x}}{{2 + 5x}}}} < \dfrac{{25}}{4}$

Answer 1

Hint: \We can say that we have to solve the problem by linear inequalities. We can see that there are two terms in the equation. So we’ll first make the value equal to both sides. If we square 5/2 it will be ${(\dfrac{5}{2})^2} = \dfrac{{25}}{4}$. Now we have to equalize both sides to compare the power. So we’ll make $\dfrac{{25}}{4}$ as $\dfrac{1}{{{{(\dfrac{2}{5})}^{^2}}}}$ .
If we simplify $\dfrac{1}{{{{(\dfrac{2}{5})}^{^2}}}}$ we’ll get ${(\dfrac{2}{5})^{ - 2}}$. Now we’ll make the power equal and find the value.

Complete answer:
First we’ll write the equation.
${(\dfrac{2}{5})^{\dfrac{{6 - 5x}}{{2 + 5x}}}} < \dfrac{{25}}{4}$
Now we’ll modify $\dfrac{{25}}{4}$ as ${(\dfrac{2}{5})^{ - 2}}$ .
${\left( {\dfrac{2}{5}} \right)^{\dfrac{{6 - 5x}}{{2 + 5x}}}} < {\left( {\dfrac{2}{5}} \right)^{ - 2}}$
So, we get, ${\left( {\dfrac{2}{5}} \right)^{\dfrac{{6 - 5x}}{{2 + 5x}}}} < {\left( {\dfrac{2}{5}} \right)^{ - 2}}$
Now, the bases are the same. So, we can compare the powers.
We know that $\left( {\dfrac{2}{5}} \right)$ is less than one. Also, we know that if we increase the power of any number less than one, the overall number decreases.
So, if ${\left( {\dfrac{2}{5}} \right)^{\dfrac{{6 - 5x}}{{2 + 5x}}}}$ is lesser than ${\left( {\dfrac{2}{5}} \right)^{ - 2}}$. This means that the power $\left( {\dfrac{{6 - 5x}}{{2 + 5x}}} \right)$is more than $ - 2$.
So, we get, $\left( {\dfrac{{6 - 5x}}{{2 + 5x}}} \right) > - 2$
Now, we solve this inequality to get the values of x.
So, adding two to both sides, we get,
$ \Rightarrow \left( {\dfrac{{6 - 5x}}{{2 + 5x}}} \right) + 2 > 0$
Taking LCM,
$ \Rightarrow \dfrac{{6 - 5x + 2\left( {2 + 5x} \right)}}{{2 + 5x}} > 0$
$ \Rightarrow \dfrac{{6 - 5x + 4 + 10x}}{{2 + 5x}} > 0$
Simplifying further,
$ \Rightarrow \dfrac{{10 + 5x}}{{2 + 5x}} > 0$
Dividing both numerator and denominator by five, we get,
$ \Rightarrow \dfrac{{x + 2}}{{\left( {x + \dfrac{2}{5}} \right)}} > 0$
Now, either both the terms $x + 2$ and $\left( {x + \dfrac{2}{5}} \right)$ must be positive or both must be negative.
So, if $x + 2 > 0$ and $\left( {x + \dfrac{2}{5}} \right) > 0$. We get, $x > - 2$ and $x > - \dfrac{2}{5}$.
So, we get the intersection as $x > - \dfrac{2}{5}$.
Also, if both are negative, we get, $x + 2 < 0$ and $x + \dfrac{2}{5} < 0$.
Hence, $x < - 2$ and $x < - \dfrac{2}{5}$.
Therefore, $x \in \left( { - \inf , - 2} \right) \cup \left( { - \dfrac{2}{5},\inf } \right)$
So this is how we have calculated the range for x .

Note:
Whenever you’ll get this kind of question always try to find the range by using linear inequality theorem. First try to equal both sides by simplifying. Then try to do as many operations as you can do to make the right side zero. When you’ll make the right side value equal to zero. Thus you’ll find the range of x with the values. Remember to make the range in between negative infinity to positive infinity.