Question

How do you solve ${(\dfrac{1}{2})^x} = 32?$

Answer 1

In order to solve this question use the “Law of indices”. In this question, law of indices for negative powers, law of indices for division and law of indices for zero power will be used. Law of indices for negative power is given as
${a^{ - b}} = \dfrac{1}{{{a^b}}}$

Law of indices for multiplication is given as
${a^b} \times {a^c} = {a^{b + c}}$

Complete step by step solution:
To solve this question, we should have the idea about the law of indices

In order to understand what is law of indices?
We have to first understand what is index or indices. Index or indices is the power from which a number is raised and hence law of indices deals with the power of numbers, that is, you can say it deals with powers for the algebraic operations of the powers and simplifying the equation easily.

Now coming to the question,
$\Rightarrow {(\dfrac{1}{2})^x} = 32$
From the law of indices for negative powers, we know that
${a^{ - b}} = \dfrac{1}{{{a^b}}}$
So writing ${(\dfrac{1}{2})^x}$ as ${2^{ - x}}$
$\Rightarrow {(\dfrac{1}{2})^x} = 32 \\\ \Rightarrow {2^{ - x}} = 32 \\\$
Again writing $32$ in terms of $2$ as

$\Rightarrow {2^{ - x}} = 32 \\\ \Rightarrow {2^{ - x}} = 2 \times 2 \times 2 \times 2 \times 2 \\\$

Now from law of indices for multiplication, we know that
${a^m} \times {a^n} = {a^{m + n}}$

Using this in our equation, we will get
$\Rightarrow {2^{ - x}} = 2 \times 2 \times 2 \times 2 \times 2 \\\ \Rightarrow {2^{ - x}} = {2^5} \\\$

Hence the base of both sides is equal, so we can equate their powers as
$\Rightarrow - x = 5 \\\ \Rightarrow x = - 5 \\\$
We got our desired solution, i.e. $x = - 5$

Note: Apart from multiplication and negative powers, there exists some more application of law of indices some of them are given below
1. For division: which is given as
${a^m} \div {a^n} = \dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
2. For zero power: which is given as
${a^0} = 1$
3. For bracket: which is given as
${({a^m})^n} = {a^{m \times n}}$
4. For fraction: which is given as
${a^{\dfrac{m}{n}}} = {(\sqrt[n]{a})^m}$