How can I make analytical demonstration, not algebraic, the next inequality log2(3)>log(4)5?

To prove the inequality above IN ANALYTICAL MANNER and not algebraic one, you have to consider the function f(x)=logx(x+1), and to prove that this function is decreasing one!

We'll write the function f(x)=ln(x+1)/ln x.

it's derivative is

f'(x)=[xln x-(x+1)ln(x+1)]/x(x+1)(ln x)^2.

As you can see, the denominator of f'(x) is positive, for any value of x>1.

For establishing the sign of denominator, we'll consider the function

g(x)=xlnx,

with the derivative

g'(x)=(ln x) +1.

For x>1, g'(x)>0, so from here we inferred that g(x) is an increasing function., so g(x)<g(x+1, so the numerator of f'(x) is negative.

Finally f'(x)<0, so f(x) is strictly decreasing for any x>1.

From here, the  inequality log2(3)>log(4)5 is true, because

3<5,

but log2(3)>log(4)5

because of the character of the function, which is decreasing one!