In a right-angled triangle, the sides are $a,\ b$ and $c$, with $c$ as hypotenuse, and $c - b
eq 1,\ c + b
eq 1$. Then the value of $(\log_{c + b}{a + \log_{c - b}a})/(2\log_{c + b}{a imes \log_{c - b}a})$ will be

Question

In a right-angled triangle, the sides are $a,\ b$ and $c$, with $c$ as hypotenuse, and $c - b 
eq 1,\ c + b 
eq 1$. Then the value of $(\log_{c + b}{a + \log_{c - b}a})/(2\log_{c + b}{a 	imes \log_{c - b}a})$ will be

Dr. Abubacker · Accepted Answer

(d)
$\frac{\log_{c + b}{a + \log_{c - a}a}}{2\log_{c + b}a.\log_{c - b}a\ }$
$= \frac{\frac{\log a}{\log{(c + b)}} + \frac{\log a}{\log{(c - b)}}}{2\frac{\log a}{\log{(c + b)}}.\frac{\log a}{\log{(c - b)}}}$
$= \frac{\log{a\{\log{(c - b) + \log{(c + b)}}\}}}{2\left( \log a \right)^{2}} = \frac{\log{(c^{2} - b^{2})}}{2\log a}$
$= \frac{\log a^{2}}{\log a^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because a^{2} + b^{2} = c^{2})$
$= 1$

Q. In a right-angled triangle, the sides are $a,\ b$ and $c$, with $c$ as hypotenuse, and $c - b \neq 1,\ c + b \neq 1$. Then the value of $(\log_{c + b}{a + \log_{c - b}a})/(2\log_{c + b}{a \times \log_{c - b}a})$ will be

Explanation

Reviewed by Dr. Abubacker

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