Question

$4608 = {4}^{a} \times {a}^{b} \times {b}^{a}$
Then find the value of -
${a}^{a} + {b}^{b}$
Don't give nonsense answers .​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:4608 = {4}^{a} \times {a}^{b} \times {b}^{a}$

Let first factorize 4608.

So,

$\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:4608 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:2304 \:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:1152\:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:576 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:288 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:144 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:72 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:36\:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:18 \:\:}} \\{\underline{\sf{3}}}& \underline{\sf{\:\:9 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:3 \:\:}} \\ \underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}$

$\rm :\longmapsto\:4608 = {2}^{9} \times {3}^{2}$

$\rm :\longmapsto\:4608 = {2}^{3 + 6} \times {3}^{2}$

$\rm :\longmapsto\:4608 = {2}^{3} \times {2}^{6} \times {3}^{2}$

$\rm :\longmapsto\:4608 = {2}^{3} \times {4}^{3} \times {3}^{2}$

can be re-arranged as

$\rm :\longmapsto\:4608 = {4}^{3} \times {3}^{2} \times {2}^{3}$

So, on comparing with

$\rm :\longmapsto\:4608 = {4}^{a} \times {a}^{b} \times {b}^{a}$

we get

$\rm \implies\:a = 3 \: \: and \: \: b = 2$

Now, Consider

$\rm :\longmapsto\: {a}^{a} + {b}^{b}$

$\rm \:  =  \: {3}^{3} + {2}^{2}$

$\rm \:  =  \: 27 + 4$

$\rm \:  =  \: 31$

Hence,

$\rm :\longmapsto\: \boxed{\tt{ {a}^{a} + {b}^{b} = 31 \: }}$