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

Asked on by sirserie

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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


but log2(3)>log(4)5

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



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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.


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