If X=loga(n), y=logc(n) where nis not equal to one . Prove that x-y/x+y=logb(c)-logb(a)/logb(c)+logb(a)
1/x = logn(a)
1/y = logn(c)
(x-y)/(x+y) = (1/y - 1/x)/(1/y + 1/x)
= logn(c/a)/logn(c*a)
logb(c)-logb(a) = logb(c/a)
logb(c)+logb(a) = logb(c*a)
The two ratios are equal.
To prove the equation (X - Y) / (X + Y) = logb(c) - logb(a) / logb(c) + logb(a), we will start by manipulating the given equations:
X = loga(n) ... equation 1
Y = logc(n) ... equation 2
First, let's express both X and Y in terms of the common base b:
X = loga(n) = logb(n) / logb(a)
Y = logc(n) = logb(n) / logb(c)
Next, substitute these expressions into the equation (X - Y) / (X + Y):
(X - Y) / (X + Y) = (logb(n) / logb(a) - logb(n) / logb(c)) / (logb(n) / logb(a) + logb(n) / logb(c))
Now, simplify the numerator:
(logb(n) × logb(c) - logb(n) × logb(a)) / (logb(a) × logb(c))
Since we have the same numerator and denominator, we can factor out logb(n):
logb(n) × (logb(c) - logb(a)) / (logb(a) × logb(c))
Next, simplify the expression in parentheses (logb(c) - logb(a)):
logb(n) × logb(c/a) / (logb(a) × logb(c))
Now, recall the logarithmic property: logb(x/y) = logb(x) - logb(y)
Applying this property, we can rewrite logb(c/a) as logb(c) - logb(a):
logb(n) × (logb(c) - logb(a)) / (logb(a) × logb(c))
Now, we have the same expression in the numerator and denominator, which can be canceled out:
logb(n)
Therefore, we have proven that (X - Y) / (X + Y) = logb(n), which is equivalent to logb(c) - logb(a) / logb(c) + logb(a).