How do you solve the expression ${4^x} \cdot {5^{4x + 3}} = {10^{2x + 3}}$?

Hint: We are given an exponential expression. We have to determine the value of x. First, apply the logarithm on both sides of the equation. Then, apply the logarithmic product property on the left-hand side of the expression. Then, apply the power property of logarithms to both sides of the equation. Then, isolate the variable terms and other terms on each side of the equation. Then, take the common factor out from the left-hand side of the equation. Then, simplify the expression and solve for x.

Complete step-by-step solution:
Given an equation, ${4^x} \cdot {5^{4x + 3}} = {10^{2x + 3}}$.
Now, we will start by taking logs on both sides of the equation.
$ \Rightarrow \log \left( {{4^x} \cdot {5^{4x + 3}}} \right) = \log \left( {{{10}^{2x + 3}}} \right)$
Now, we will apply the product rule of logarithms ${\log _a}\left( {bc} \right) = {\log _a}\left( b \right) + {\log _a}\left( c \right)$ to the left hand side of the equation.
$ \Rightarrow \log \left( {{4^x}} \right) + \log \left( {{5^{4x + 3}}} \right) = \log \left( {{{10}^{2x + 3}}} \right)$
Now, we will apply the power rule of logarithms ${\log _a}\left( {{b^c}} \right) = c {\log _a}\left( b \right)$ to both sides of the equation.
$ \Rightarrow x\log \left( 4 \right) + \left( {4x + 3} \right)\log \left( 5 \right) = \left( {2x + 3} \right)\log \left( {10} \right)$
Now, we will apply the distributive property at both sides of the equation.
$ \Rightarrow x\log \left( 4 \right) + 4x\log \left( 5 \right) + 3\log \left( 5 \right) = 2x\log \left( {10} \right) + 3\log \left( {10} \right)$
Move all variable terms on the left hand side of the equation and without variables on the right hand side of the equation.
$ \Rightarrow x\log \left( 4 \right) + 4x\log \left( 5 \right) - 2x\log \left( {10} \right) = 3\log \left( {10} \right) - 3\log \left( 5 \right)$
Now, factor out x from the left hand side of the equation.
$ \Rightarrow x\left[ {\log \left( 4 \right) + 4\log \left( 5 \right) - 2\log \left( {10} \right)} \right] = 3\log \left( {10} \right) - 3\log \left( 5 \right)$
Now, solve for x.
$ \Rightarrow x = \dfrac{{3\log \left( {10} \right) - 3\log \left( 5 \right)}}{{\log \left( 4 \right) + 4\log \left( 5 \right) - 2\log \left( {10} \right)}}$
Now, we will substitute $\log \left( {10} \right) = 1$, $\log \left( 5 \right) = 0.698$ and $\log \left( 4 \right) = 0.602$ into the expression.
$ \Rightarrow x = \dfrac{{3 \times 1 - 3 \times 0.698}}{{0.602 + 4 \times 0.698 - 2 \times 1}}$
On simplifying the expression, we get:
$ \Rightarrow x = \dfrac{{3 - 2.094}}{{0.602 + 2.792 - 2}}$
On further simplifying the expression, we get:
$ \Rightarrow x = \dfrac{{0.906}}{{1.394}}$
$ \Rightarrow x \approx 0.65$
Hence the value of x for the expression ${4^x} \cdot {5^{4x + 3}} = {10^{2x + 3}}$ is approximately $0.65$.

Note: The students must note that in such types of questions various logarithmic properties are applied. The logarithmic properties are as follows:
The product rule of logarithms
$ \Rightarrow {\log _a}\left( {bc} \right) = {\log _a}\left( b \right) + {\log _a}\left( c \right)$
The quotient rule of logarithms
$ \Rightarrow {\log _a}\left( {\dfrac{b}{c}} \right) = {\log _a}\left( b \right) - {\log _a}\left( c \right)$
The power rule of logarithms
$ \Rightarrow {\log _a}\left( {{b^c}} \right) = c{\log _a}\left( b \right)$
The zero exponent rule of logarithms
$ \Rightarrow {\log _a}\left( 1 \right) = 0$
There is an alternative way to solve the problem which is to simplify the given expression where we make the bases both side same and then use the exponent property.