How do you solve ${\log _3}\left( {4x - 5} \right) = 5$?

Hint: We will use some identities and formulas of logarithmic functions and then, use them on the given expression, we will then obtain the required answer that is the value of x.

Complete step-by-step answer:
We are given that we need to solve ${\log _3}\left( {4x - 5} \right) = 5$.
For this, we will first of all use: ${\log _b}a = \dfrac{{\log a}}{{\log b}}$.
On replacing a by 4x – 5 and b by 3, we will then obtain:-
$ \Rightarrow {\log _3}\left( {4x - 5} \right) = \dfrac{{\log \left( {4x - 5} \right)}}{{\log 3}}$
Putting this with the given right hand side, we will then obtain:-
$ \Rightarrow \dfrac{{\log \left( {4x - 5} \right)}}{{\log 3}} = 5$
On taking the log 3 from division in the denominator of left hand side to multiplication in the numerator of right hand side, we will then obtain the following expression:-
$ \Rightarrow \log \left( {4x - 5} \right) = 5 \times \log 3$ …………….(1)
Now, we know that: $\log {a^b} = b \times \log a$
Replacing a by 3 and b by 5, we will then obtain the following expression:-
$ \Rightarrow 5 \times \log 3 = \log {3^5}$
Putting this in equation number (1), we will then obtain the following expression:-
$ \Rightarrow \log \left( {4x - 5} \right) = \log {3^5}$
Now, we can cancel out log from both the sides to obtain the following expression:-
$ \Rightarrow 4x - 5 = {3^5}$ ……………………(2)
We also know that ${3^5} = 3 \times 3 \times 3 \times 3 \times 3$
On solving it, we will then obtain: ${3^5} = 243$
Putting this in equation number (2) to obtain the following expression:-
$ \Rightarrow 4x - 5 = 243$
Taking the 5 from subtraction in the left hand side to addition in the right hand side, we will then obtain the following expression:-
$ \Rightarrow 4x = 5 + 243$
Simplifying the calculation on right hand side by adding the required quantities to obtain the following expression:-
$ \Rightarrow $ 4x = 248
Dividing both sides by 4 to get the final result written as follows:-
$ \Rightarrow $ x = 62

Hence, the value of x is 62.

Note:
The students must note that when we cut off log from both sides, it is due to a fact hidden beside it. Let us understand it:-
We know that if log a = b, then we have $a = {e^b}$.
Therefore, if we are given that log a = log b, then we get: $a = \log {e^b}$
Now we will use the fact that $\log {a^b} = b \times \log a$.
Therefore, we get: $a = b\log e$.
We know that log e = 1.
Therefore, we have a = b and thus we have the required formula which we used in the above solution.