How do you solve the expression ${9^{4x + 1}} = 64$?

Hint: In this question we have to solve the given equation which is in exponent form, we will solve the given question by using properties of both logarithms, first take logarithms on both sides of the equation and then solve the equation to get the required result.

Complete step-by-step solution:
A logarithm is an exponent which indicates to what power a base must be raised to produce a given number.
 $y = {b^x}$exponential form,
$x = {\log _b}y$ logarithmic function, where $x$ is the logarithm of $y$ to the base $b$, and ${\log _b}y$ is the power to which we have to raise $b$ to get $y$, we are expressing $x$ in terms of $y$.
The given equation is ${9^{4x + 1}} = 64$,
Now taking logarithms on both sides of the equation we get,
$ \Rightarrow \log {9^{4x + 1}} = \log 64$,
Now using logarithmic identity ${\log _a}{x^n} = n{\log _a}x$ on the left side of the equation we get,
$ \Rightarrow \left( {4x + 1} \right)\log 9 = \log 64$,
Now using distributive property we get,
$ \Rightarrow 4x\log 9 + \log 9 = \log 64$,
Now subtract from both sides of the equation with $\log 9$, we get,
$ \Rightarrow 4x\log 9 + \log 9 - \log 9 = \log 64 - \log 9$,
Now simplifying we get,
$ \Rightarrow 4x\log 9 = \log 64 - \log 9$,
Now divide both sides of the equation with $\log 9$, we get,
$ \Rightarrow \dfrac{{4x\log 9}}{{\log 9}} = \dfrac{{\log 64 - \log 9}}{{\log 9}}$,
Now simplifying we get,
$ \Rightarrow 4x = \dfrac{{\log 64 - \log 9}}{{\log 9}}$,
Now using logarithmic table we get,
$ \Rightarrow 4x = \dfrac{{1.806 - 0.954}}{{0.954}}$,
Now simplifying we get,
$ \Rightarrow 4x = \dfrac{{0.8517}}{{0.954}}$,
Now again simplifying we get,
$ \Rightarrow 4x = 0.892$,
Now divide both sides with 4 we get,
$ \Rightarrow \dfrac{{4x}}{4} = \dfrac{{0.892}}{4}$,
Now simplifying we get,
$ \Rightarrow x = 0.223$,
The value of $x$ is equal to $0.223$.

$\therefore $The value of $x$ is equal to $0.223$ when ${9^{4x + 1}} = 64$.

Note: A logarithm is a mathematical operation that determines how many times a certain number, called the base, is multiplied by itself to reach another number, in these types of questions, we use logarithmic properties and formulas, and some of useful formulas are:
 ${\log _a}xy = {\log _a}x + {\log _a}y$,
${\log _a}{x^n} = n{\log _a}x$,
${\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}$,
${\log _{\dfrac{1}{a}}}b = - {\log _a}b$,
${\log _a}a = 1$,
${\log _{{a^x}}}b = \dfrac{1}{x}{\log _a}b$.