Question

Evaluate the logarithmic function ${\log _2}128 + {\log _3}243$

Answer 1

Hint : For a logarithmic function the following formulas hold true.
${\log _a}{a^x} = x{\log _a}a$ and ${\log _a}a = 1$
So solve the given logarithms in order to bring them to this form and solve accordingly using these formulas.

Complete step-by-step answer :
Given to us a logarithmic function ${\log _2}128 + {\log _3}243$
In order to solve this, let us find the value of each logarithm separately.
So first, let us solve ${\log _2}128$
Let us write $128$ as ${2^x}$ and in order to find the value of x, we need to prime factorize $128$
From this we can write the value of
$128$ as $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ which is ${2^7}$
Now we equate ${2^x} = {2^7}$ and hence the value of x is seven i.e. $x = 7$
Now the logarithm can be written as ${\log _2}{2^7}$ , but we already know that
${\log _a}{a^x} = x{\log _a}a$ and hence this can be written as follows
${\log _2}{2^7} = 7{\log _2}2$
We also know that ${\log _a}a = 1$ and hence the value of ${\log _2}2$ is one.
Hence the value of ${\log _2}128 = 7$
Similarly we can find the value of ${\log _3}243$
We write the value of $243$ as ${3^x}$ and we use prime factorization to find the value of x.
Hence, the value of $243$ can be written as
$3 \times 3 \times 3 \times 3 \times 3$ which is ${3^5}$
So the value of x now becomes five.
We substitute this in the logarithm as ${\log _3}{3^5}$ which can also be written as
$5{\log _3}3$
Therefore the value of ${\log _3}243$ is $5$ since from the formula ${\log _a}a = 1$
Now, we add the values of both the logarithms as follows
${\log _2}128 + {\log _3}243 = 7 + 5 = 12$
So, the correct answer is “12”.

Note : It is to be noted that for a logarithmic function ${\log _a}b$ to be in the form of ${\log _a}{a^x}$ and this logarithmic value to be x, the number b should be a multiple of a. In this case $128$ is a multiple of $2$ and $243$ is a multiple of $3$