How do you solve \[{16^{5x}} = {64^{x + 7}}\]?

Hint: Here, we will rewrite the base on both sides using exponents such that our base is the same. We will then equate the powers and form a linear equation. Then by solving it, we will be able to find the required value of \[x\] and hence the required answer.

Formula Used:
 \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\]

Complete step by step solution:
In order to solve this question, we will make the base of the given exponents the same.
We can write 16 and 64 as:
\[16 = 4 \times 4 = {4^2}\]
\[64 = 4 \times 4 \times 4 = {4^3}\]
Now, we will substitute these values in the given expression.
Thus, we get,
\[{\left( {{4^2}} \right)^{5x}} = {\left( {{4^3}} \right)^{x + 7}}\]
Now, using the exponential properties \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\], we get
\[ \Rightarrow {4^{10x}} = {4^{3x + 21}}\]
Now, since the base of both the powers is the same, we can equate the powers. Therefore, we get
\[10x = 3x + 21\]
Subtracting \[3x\] from both the sides, we get
\[ \Rightarrow 7x = 21\]
Dividing both sides by 7, we get
\[ \Rightarrow x = 3\]

Therefore, the required value of \[x\] is 3.

Additional information:
An expression that represents the repeated multiplication of the same number is known as power. Whereas, when a number is written with power then the power becomes the exponent of that particular number. It shows how many times that particular number will be multiplied by itself. Hence, whenever we are given the multiplication of the same numbers, then we can express that number with an exponent and vice-versa.

Note:
In this question, we can also check whether our answer is correct or not by substituting our answer \[x = 3\] in \[{16^{5x}} = {64^{x + 7}}\]
Thus, we get,
\[ \Rightarrow {16^{5\left( 3 \right)}} = {64^{3 + 7}}\]
\[ \Rightarrow {16^{15}} = {64^{10}}\]
This can again be written as:
\[ \Rightarrow {\left( {{4^2}} \right)^{15}} = {\left( {{4^3}} \right)^{10}}\]
Here, using \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\], we get,
\[ \Rightarrow {4^{30}} = {4^{30}}\]
Clearly, LHS \[ = \] RHS
Hence, it is verified that our answer is correct.