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How do you solve ${4^{ - 3x}} = 0.25$?

Last updated date: 19th Jul 2024
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Hint: In this question, we will solve by using exponential identities, first convert the right hand side into fraction, and then use the identity $\dfrac{1}{{{a^m}}} = {a^{ - m}}$, and by using the fact that when bases are equal then equate the powers, then equate the left hand side to the right hand side and then solve for $x$ to get the required result.

Complete step by step solution:
Exponents are defined as when an expression or a statement of specific natural numbers are represented as a repeated power by multiplication of its units then the resulting number is called as an exponent. The resulting set of numbers are the same as the original sequence.
Given expression is ${4^{ - 3x}} = 0.25$ ,
Now convert the decimal to fraction on the right hand side, we get,
$ \Rightarrow {4^{ - 3x}} = \dfrac{{25}}{{100}}$,
Now simplifying we get,
$ \Rightarrow {4^{ - 3x}} = \dfrac{1}{4}$,
Now using the exponential identity, $\dfrac{1}{{{a^m}}} = {a^{ - m}}$, we get,
Here in the term on the left hand side, $a = 4$ and $m = - 3x$, now substituting these values in the identity we get,
$ \Rightarrow \dfrac{1}{{{4^{3x}}}} = \dfrac{1}{4}$,
Now as the bases are equal then equate the powers, we get,
$ \Rightarrow 3x = 1$,
Now divide both sides of the equation with 3, we get,
$ \Rightarrow \dfrac{{3x}}{3} = \dfrac{1}{3}$,
Now simplifying the expression we get,
$ \Rightarrow x = \dfrac{1}{3}$,
So the value of $x$ when the equation is solved is $\dfrac{1}{3}$.

$\therefore $ The value of $x$ when the given equation i.e., ${4^{ - 3x}} = 0.25$ is solved is $\dfrac{1}{3}$.

Note: There are various laws of exponents we should remember and practise in order to solve and understand the exponential concept. The following are some of the exponent laws:
${a^0} = 1$,
${a^m} \times {a^n} = {a^{m + n}}$,
$\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$,
$\dfrac{1}{{{a^m}}} = {a^{ - m}}$,
${a^m} \times {b^m} = {\left( {ab} \right)^m}$,
$\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}$.