Find the value of ${2^2}^{ + 2{{\log }_2}3}$

Hint: We know that ${\log _a}a$ is equal to $1$. By using this we will add in the power of $2$ with $2$ as making $2{\log _2}2$. Thereafter we will add the log’s value in the power of $2$. Further we will use ${a^{{{\log }_a}x}} = x$ to find the value.
By using some exponent formula:
(i) ${a^{m + n}} = {a^m} \times {a^n}$
(ii) ${a^{{{\log }_a}x}} = x$


Complete step by step solution:
Here, ${2^2}^{ + {{\log }_2}3}$
As we know that ${\log _2}2 = 1$
$ = {2^{2{{\log }_2}2}}^{ + {{\log }_2}3}$
$ = {2^{{{\log }_2}{{(2)}^2}}}^{ + {{\log }_2}3}$ $[m\log a = \log {a^m}]$
$ = {2^{{{\log }_2}4}}^{ + {{\log }_2}3}$
$ = {2^{{{\log }_2}4 \times 3}}$ $[\log m + \log n = \log m \times n]$
$ = {2^{{{\log }_2}12}}$
Using $[{a^{{{\log }_a}x}} = x]$in the above equation, we have
$ \Rightarrow 12$
Therefore,${2^{2 + {{\log }_2}3}} = 12$
Additional Information: Some logarithm formulae are:
(i) ${\log _x}(a \times b) = {\log _x}a + {\log _x}b$
(ii) ${\log _x}\left( {\dfrac{a}{b}} \right) = {\log _x}a - {\log _x}b$
(iii) ${\log _x}({a^b}) = b\,{\log _x}a$
(iv) ${\log _x}(1) = 0$
(v) ${\log _x}({b^k}) = k$
(vi) ${b^{{{\log }_b}}}^{(k)} = k$


Note: In this type of question students should convert the whole value in log form whether it is in base or in power, so use the formula to get the correct answer.