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
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
because of the character of the function, which is decreasing one!
To prove analytically log2(3) > log4(5) where, left side expresion is the logarithm of 3 with respect to base 2, and the right is the logarithm of 5 with respect to base 4.
We take the left side log2(4) and try what is the expression value in terms of logarthim with respect to base 4 and then we can easily compare the value with the expression on the right side and say about the veracity of the given inequality.
log2(3) = log3/log2. Here,in the right logarithm, I did not say the base, but the base could be any thing but should be same to both numerator and denominator.
Multiply by 2 both numerator and denominator.
Thus lo2(3) is equal to log4(9) which could be compared to the right side expression , which is log4(5). So,
log4(9) > log4(5) or the given inequality ,
log2(3) > log4(5) is proved to hold good.