If a^{2}+b^{2}=c^{2}, then frac{1}{log _{c+a} | vedclass

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

Question

(B) Given the expression $\frac{1}{\log _{c+a} b}+\frac{1}{\log _{c-a} b}$.Using the change of base property $\frac{1}{\log _{x} y} = \log _{y} x$,we

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given the expression $\frac{1}{\log _{c+a} b}+\frac{1}{\log _{c-a} b}$.Using the change of base property $\frac{1}{\log _{x} y} = \log _{y} x$,we can rewrite the expression as:$\log _{b}(c+a) + \log _{b}(c-a)$.Using the logarithmic property $\log _{m} n + \log _{m} p = \log _{m} (n \cdot p)$,we get:$\log _{b}((c+a)(c-a)) = \log _{b}(c^{2}-a^{2})$.Since $a^{2}+b^{2}=c^{2}$,it follows that $c^{2}-a^{2}=b^{2}$.Substituting this into the expression:$\log _{b}(b^{2}) = 2 \log _{b} b = 2(1) = 2$.

If $a^{2}+b^{2}=c^{2},$ then $\frac{1}{\log _{c+a} b}+\frac{1}{\log _{c-a} b}=$

Similar Questions

If $0 < a \leq x$,the minimum value of $\log_{a} x + \log_{x} a$ is

If $0 < a < 1$,$0 < x < 1$,and $x < a$,then $\log_{a} x$ is:

$x^{\log y - \log z} \cdot y^{\log z - \log x} \cdot z^{\log x - \log y} = ?$

$\frac{\log _{9} 11}{\log _{5} 13}-\frac{\log _{3} 11}{\log _{\sqrt{5}} 13}=$

If $\log _{30} 3=a$ and $\log _{30} 5=b,$ then find the value of $\log _{30} 8.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module