Question

How do you evaluate ${\log _8}2$?

Answer 1

Hint: In questions like this, we can assume the value which needs to be calculated as $x$ so that we can convert the equation from logarithmic form to exponential form and then we need to find a relation between LHS and RHS so that it makes a linear equation through which we can find the value of the variable $x$.

Complete step by step solution:
(i) We are given to calculate,
${\log _8}2$
Let us assume ${\log _8}2$ as $x$
$x = {\log _8}2$
Since we know that if $a$ and $b$ are positive real numbers and $b$ is not equal to $1$, then ${\log _b}a = y$ is equivalent to ${b^y} = a$.
Using the above-mentioned property here, we will get:
${8^x} = 2$

(ii) Now, since we have to simplify the equation to solve it, we will try to make the base equal on both the sides. So, as we know that ${2^3} = 8$, we will substitute $8$ as ${2^3}$ in the above equation so that the base is equal on both the sides.
${({2^3})^x} = 2$

(iii) As we know that ${({a^m})^n} = {a^{mn}}$, we can write ${({2^3})^x}$ as ${2^{3x}}$. Therefore, our equation will become:
${2^{3x}} = 2$
Also, as we know that $2 = {2^1}$. Therefore, we can write:
${2^{3x}} = {2^1}$

(iv) Since the bases on both sides are the same, we can directly equate the exponential powers on them. So, our equation becomes:
$3x = 1$
Solving the equation, we will shift $3$ on RHS, we will get”
$x = \dfrac{1}{3}$
i.e.,
$x = 0.333$
Hence, ${\log _8}2 = 0.333$

Note: Another method to solve this question is by using the property ${\log _b}a = \dfrac{1}{{{{\log }_a}b}}$. As we would have got $\dfrac{1}{{{{\log }_2}8}}$ then, we would directly write $\dfrac{1}{3}$ as the answer as we know ${\log _2}8 = 3$ since ${2^3} = 8$. Also, the most important step here is step (ii) where we have to figure out how to relate LHS and RHS so that it becomes easier to equate the exponents and obtain the value of variable $x$. In most of the questions, numbers are given in such a way that they can be expressed as the exponents of the same digit, so that the base at LHS and RHS gets equal and we can easily equate their powers.