Question

If \(x\), y, z are the sides of a right angled triangle, where z is the largest side, then \(\frac{1}{log_{x+z}y}+\frac{1}{log_{z-x}y} =\)

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Use 1/log_a b = log_b a to combine the two terms, then Pythagoras.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The change of base rule says \(\dfrac{1}{\log_a y} = \log_y a\). Also, the sum of two logs to the same base is the log of the product.

Step 2: Rewrite the expression
\[ \frac{1}{\log_{x+z} y} + \frac{1}{\log_{z-x} y} = \log_y (x+z) + \log_y (z-x) = \log_y\big[(z+x)(z-x)\big] \]

Step 3: Use the right triangle
Since z is the hypotenuse, \(x^2 + y^2 = z^2\), so \(z^2 - x^2 = y^2\).
\[ \log_y (y^2) = 2 \log_y y = 2 \]
Values 1, 3 and 4 would need the product to be \(y\), \(y^3\) or \(y^4\), which does not follow from Pythagoras.

Final Answer:
The value is 2, option (B). \[ \boxed{2} \]