If x, y, z are the sides of a right-angled tr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given that $x, y, z$ are sides of a right-angled triangle with $z$ as the hypotenuse, by Pythagoras theorem, we have $x^2 + y^2 = z^2$.This implie

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given that $x, y, z$ are sides of a right-angled triangle with $z$ as the hypotenuse, by Pythagoras theorem, we have $x^2 + y^2 = z^2$.This implies $y^2 = z^2 - x^2$, which can be factored as $y^2 = (z-x)(z+x)$.Taking the logarithm with base $y$ on both sides: $\log_y(y^2) = \log_y((z-x)(z+x))$.$2 = \log_y(z-x) + \log_y(z+x)$.Using the change of base formula $\log_a b = \frac{1}{\log_b a}$, we have $\frac{1}{\log_{z-x} y} = \log_y(z-x)$ and $\frac{1}{\log_{z+x} y} = \log_y(z+x)$.Substituting these into the expression: $\frac{1}{\log_{x+z} y} + \frac{1}{\log_{z-x} y} = \log_y(z+x) + \log_y(z-x) = \log_y((z+x)(z-x)) = \log_y(y^2) = 2$.

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

Similar Questions

If $\log _{2} 6 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8)$, then the values of $x$ are

The number of digits in $20^{301}$ (given, $\log _{10} 2=0.3010$) is:

If $m^n = n^m$,then the value of $m$ in terms of $n$ is:

If $\frac{1}{2} \le \log_{0.1} x \le 2$,then:

If $x = \log_3 5$ and $y = \log_{17} 25$,then which of the following is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module