Logarithm identity

$\begin{aligned} \frac{2+\log_2{3}}{1+\log_2{3}} + \frac{3+\log_3{4}} {1+\log_3{2}} & = \frac{2+\frac{\log{3}}{\log{2}}}{1+\frac{\log{3}}{\log{2}}} + \frac{3+\frac{2\log{2}}{\log{3}}}{1+\frac{\log{2}}{\log{3}}} \\ & = \frac{2\log{2} + \log{3}}{\log{2} + \log{3}} + \frac{3\log{3} + 2\log{2}}{\log{3} + \log{2}} \\ & = \frac{4\log{2} + 4\log{3}}{\log{2} + \log{3}} \\ & = \boxed{4} \end{aligned}$

$\frac { 2+\log _{ 2 }{ 3 } }{ 1+\log _{ 2 }{ 3 } } +\frac { 3+\log _{ 3 }{ 4 } }{ 1+\log _{ 3 }{ 2 } } \\ =\frac { \log _{ 2 }{ 4 } +\log _{ 2 }{ 3 } }{ \log _{ 2 }{ 2 } +\log _{ 2 }{ 3 } } +\frac { \log _{ 3 }{ 27 } +\log _{ 3 }{ 4 } }{ \log _{ 3 }{ 3 } +\log _{ 3 }{ 2 } } \\ =\frac { \log _{ 2 }{ 12 } }{ \log _{ 2 }{ 6 } } +\frac { \log _{ 3 }{ (27\times 4) } }{ \log _{ 3 }{ 6 } } \\ =\log _{ 6 }{ 12 } +\log _{ 6 }{ 108 } \\ =\log _{ 6 }{ (27\times 4\times 12) } \\ =\log _{ 6 }{ (3^{ 4 }\cdot { 2 }^{ 4 }) } \\ =4\cdot \log _{ 6 }{ 6 } \\ =4$

Answer is $4$ .

We have:

$\quad\dfrac{2+\log_23}{1+\log_23}+\dfrac{3+\log_34}{1+\log_32}\\ =\dfrac{2+\log_23}{1+\log_23}+\dfrac{3+2\log_32}{1+\log_32}\\ =1+\dfrac{1}{1+\log_23}+2+\dfrac{1}{1+\log_32}\\ =3+\dfrac{1+\log_32+1+\log_23}{(1+\log_23)(1+\log_32)}\\ =3+\dfrac{2+\log_32+\log_23}{1+\log_23+\log_32+\log_23\log_32}\\ =3+\dfrac{2+\log_32+\log_23}{2+\log_23+\log_32}=\boxed{4}$

e simplification can be done in various methods but the best method is to separate it as far as possible - like 2/1+log 2(3) + log 2(3)/1+log 2(3) + .... so on. I've done it in my own method and got it as 1 + log 6(216) = 4

2 + log 2(3) = log 2(4) + log 2(3) = log 2(12) 1 + log 2(3) = log 2(2) + log 2(3) = log 2(6) 3 + log 3(4) = log 3(27) + log 3(4) = log 3(108) 1 + log 3(2) = log 3(3) + log 3(2) = log 3(6)

Now use the change of base formula.

log 2(12)/log 2(6) = log 6(12) log 3(108)/log 3(6) = log 6(108)

So now the original expression can be written as follows.

log 6(12) + log 6(108) = log 6(1296) = log 6(6^4) = 4

An easy way to simplify the solution is see that if you make log(base 2) of 3 = x then log(base 3) of 2 = 1 / x and work accordingly. Grace and peace