Question

If a2 + b2 = c2 then
Loga (c-b) + Loga (c + b) =
A)-1 B)-2
C)1
D) 2​

Answer 1

$log(c^{2}-b ^{2} )_{a}$$log(c-b)_{a} + log(c+b)_{a}$  =  2

Given:

$a^{2}+ b^{2}= c^{2}$

To Find:

$log(c-b)_{a} + log(c+b)_{a}$  =  ?

Solution:

According to the property of log:

$log(x)_{a} + log(y)_{a}=log(xy)_{a}$  

Similarly;

$log(c-b)_{a} + log(c+b)_{a}$  

=> $log((c-b)(c+b))_{a}$

=> $log(c^{2}-b ^{2} )_{a}$

=> $log(a)^{2}_{a}$

=> 2$loga_{a}$

According to properly of log:

$logx_{x}$ = 1

Similarly;

2$loga_{a}$ = 2

Hence, $log(c^{2}-b ^{2} )_{a}$$log(c-b)_{a} + log(c+b)_{a}$  =  2

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