How do you solve ${8^x} = 4$ ?

The given equation is ${8^x} = 4$

We can write 4 as the product of 2 and 2 i.e.

$ \Rightarrow 4 = 2 \times 2$

$ \Rightarrow 4 = {2^2}$

Also, we can write 8 as the product of 2 and 2 and 2 i.e.

$ \Rightarrow 8 = 2 \times 2 \times 2$

$ \Rightarrow 8 = {2^3}$

Now putting values of 8 and 4 in the given equation i.e. ${8^x} = 4$

$ \Rightarrow {\left( {{2^3}} \right)^x} = {2^2}$

$ \Rightarrow {2^{3x}} = {2^2}$

Now, taking ${\log _2}\left( x \right)$ both sides of the above equation

$ \Rightarrow {\log _2}\left( {{2^{3x}}} \right) = {\log _2}\left( {{2^2}} \right)$

$ \Rightarrow 3x{\log _2}\left( 2 \right) = 2{\log _2}\left( 2 \right)$

As we know that ${\log _2}\left( 2 \right) = 1$ , hence we left with

$ \Rightarrow 3x = 2$

Now, dividing both sides by 3, we get

$ \Rightarrow x = \dfrac{2}{3}$

If you have ${a^x} = {a^y}$ , where a is a constant value and x and y are variable, then you can directly write it as $x = y$.

Whenever you are taking the logarithm, take the log with the base of that number that is present on both sides so that it becomes 1. As in the above solution, we were having 2 on both sides so we took a logarithm with base 2 on both sides.