How do you solve ${{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( \dfrac{1}{2} \right)}^{x}}$ ?

Hint: We have been given exponential functions in fractional format in which variable-x as the powers of constant terms on the right hand side as well as the left hand side. In order to solve this, we shall use certain exponential properties. Thus, we shall convert the bases of the exponential terms on both the sides equal to $\dfrac{1}{2}$ and take logarithms on both sides to cancel out terms and obtain the value of x.

Complete step by step solution:
One of the reasons we use multiplication is because it acts as a shorthand for successive addition. Similarly, exponents act as a shorthand for successive multiplication.
Given that, ${{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( \dfrac{1}{2} \right)}^{x}}$
We have to solve the given expression and find the value of variable x. We shall start by making the bases of both the sides equal to $\dfrac{1}{2}$ as $4=2\times 2$, that is, $4={{2}^{2}}$.
Thus, ${{\left( \dfrac{{{1}^{2}}}{{{2}^{2}}} \right)}^{2x}}={{\left( \dfrac{1}{2} \right)}^{x}}$ because we know that $1\times 1=1$ and thus, ${{1}^{2}}=1$
Taking the power 2 common on the left hand side, we get
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{2\left( 2x \right)}}={{\left( \dfrac{1}{2} \right)}^{x}}$
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{4x}}={{\left( \dfrac{1}{2} \right)}^{x}}$
Taking logarithm on both sides and using the property of log, $\log {{a}^{b}}=b\log a$, we get
 $\Rightarrow \log {{\left( \dfrac{1}{2} \right)}^{4x}}=\log {{\left( \dfrac{1}{2} \right)}^{x}}$
$\Rightarrow 4x\log \dfrac{1}{2}=x\log \dfrac{1}{2}$
Dividing both sides by $\log \dfrac{1}{2}$ , we get
$\Rightarrow 4x=x$
$\Rightarrow x=0$

Therefore, the solution of ${{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( \dfrac{1}{2} \right)}^{x}}$ is $x=0$.

Note: Exponents have their own set of rules and properties according to which they can be manipulated. One of them is that if exponential terms are being multiplied or divided, then their respective powers are added or subtracted respectively. That is, ${{x}^{a}}.{{x}^{b}}={{x}^{a+b}}$ and $\dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}$ , where a and b are the powers of the base x in the exponential functions.