How do you write \[{4^{x + 1}} = 32\] in log form?

Hint: Take the \[\log \] on both the sides of the equation and then simplify the constant term to reduce the equation into its shortest form.

Formula used:
If we take the \[\log \] on both the sides of an equation, then the power on that variable can be written as the multiplication of the \[\log \] of that variable.
Let's say, \[{m^n} = c\].
If we take \[\log \]on both the sides, we can write it as following:
\[\log ({m^n}) = \log (c)\].
Now, as per the power rule of logarithm, we can derive the following equation:
\[n \times \log (m) = \log (c)\].
Division rule:
If we divide two \[\log \] values on by the other, then we can write it as following:
Let say, we are dividing \[\log (c)\] by \[\log (m)\], we can write the following expression:
\[\log \left( {\dfrac{c}{m}} \right)\].
So, division rule states that:
\[\log \left( {\dfrac{c}{m}} \right) = \log (c) - \log (m).\]

Complete step by step answer:
Now, the given question is as following:
\[ \Rightarrow {4^{x + 1}} = 32\].
Now, taking the \[\log \]on both the sides, we get:
\[ \Rightarrow \log \left( {{4^{x + 1}}} \right) = \log (32)\].
Now, by applying the power rule of logarithm, we get:
\[ \Rightarrow (x + 1) \times \log \left( 4 \right) = \log (32)\].
Now, convert \[4\]and \[32\]into power of \[2\], we get:
\[ \Rightarrow (x + 1) \times \log \left( {{2^2}} \right) = \log ({2^5})\].
Now, multiply the terms in L.H.S, we get:
\[ \Rightarrow x\log 4 + \log 4 = \log 32\].
Now, take the logarithm term in R.H.S, we get:
\[ \Rightarrow x\log 4 = \log 32 - \log 4\].
Now, by applying the division rule of logarithm, we get:
\[ \Rightarrow x\log 4 = \log \left( {\dfrac{{32}}{4}} \right)\].
Now, by doing further simplification, we get:
\[ \Rightarrow x\log 4 = \log \left( 8 \right)\].
Now, convert \[8\]into \[2's\]power, we get:
\[ \Rightarrow x\log 4 = \log \left( {{2^3}} \right)\]
Now, by applying the power rule of logarithm, we get:
\[ \Rightarrow x\log 4 = 3\log \left( 2 \right)\].

So, the final form logarithm of the given equation is \[x\log 4 = 3\log \left( 2 \right)\].

Note: We can solve the given equation for \[x\]and then we can put this value into the derived equation to cross check the solution.
The equation is as follows: \[{4^{x + 1}} = 32\].
Now, convert both the terms in the equation into \[2's\]power, we get:
\[ \Rightarrow {2^{2(x + 1)}} = {2^5}\].
By comparing both the equation, we can write the following equation:
\[ \Rightarrow 2(x + 1) = 5\].
By doing simplification, we get:
\[ \Rightarrow 2x + 2 = 5\].
Taking the constant term into R.H.S, we get:
\[ \Rightarrow 2x = 5 - 2\].
By doing the simplification, we get:
\[ \Rightarrow x = \dfrac{3}{2}\].
Now, the solved equation is : \[x.\log 4 = 3\log \left( 2 \right)\].
If we put \[x = \dfrac{3}{2}\]into the above solution, we get:
\[ \Rightarrow \dfrac{3}{2}\log 4 = 3\log \left( 2 \right)\]
By doing further simplification, we get:
\[ \Rightarrow \dfrac{3}{2}\log {2^2} = 3\log \left( 2 \right)\].
Now, applying the power rule:
\[ \Rightarrow \dfrac{3}{2} \times 2\log 2 = 3\log \left( 2 \right)\].
Now, doing further simplification, we get:
\[ \Rightarrow 3\log (2) = 3\log \left( 2 \right)\].
So, it is clear that the measure of both values in the R.H.S and L.H.S are equal.
So, the derived equation is correct.