Question

How do you write ${\log _2}32 = 5$ in exponential form?

Answer 1

Hint: Write $32\;$ in an exponential form that is writing it in the powers of $2$ . We are writing it in the powers of $2$ because it gets canceled with the base and is easier to then evaluate. By using the law of powers of logarithms evaluate further to prove LHS=RHS.

Formula used: The logarithm of a given constant $x$ is the exponent to which another fixed constant, the base $b$ , must be raised, to produce that constant $x$.
$lo{g_b}({b^x}) = x$
Law of powers of logarithms, if we have the function, $f(x) = {\log _a}({b^c})$ .Then we can convert into power form as, $f(x) = c{\log _a}(b)$ .

Complete step-by-step solution:
Firstly write $32\;$ in exponential form
Write it in powers of $2$ since it will be easy to evaluate as it gets canceled by the base.
$\Rightarrow 32 = 2 \times 2 \times 2 \times 2 \times 2$
Which can also be written as,
$\Rightarrow 32 = {2^5}$
Now write it back in the logarithmic expression.
$\Rightarrow {\log _2}{2^5} = 5$
By using the law of powers of logarithms,
If we have the function, $f(x) = {\log _a}({b^c})$ .Then we can convert into power form as, $f(x) = c{\log _a}(b)$ .
Here, $a = 2;b = 2;c = 5$
$\Rightarrow {\log _2}{2^5} = 5{\log _2}2$
Since ${\log _2}2 = 1$,
We can write it again as,
$\Rightarrow 1 \times 5 = 5$
$\Rightarrow 5 = 5$
LHS=RHS.
Now,
The logarithm of a given constant $x$ is the exponent to which another fixed constant, the base $b$ , must be raised, to produce that constant $x$.
$\Rightarrow {\log _b}({b^x}) = x$
Therefore ${\log _2}32$ can be written in exponential form as ${2^5} = 32$

Note: One should ensure that the base of the given logarithms is the same before evaluating the expression using the laws of the logarithms. While using the subtraction law of logarithms ensure which term is written in the numerator and which in the denominator. Always the first term will be in the numerator whereas the second term in the denominator.