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

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 $