1/1+loga *bc +1/1+logb*ca +1/1+logc*ab
Answers
Answer:
1/1+loga *bc +1/1+logb*ca +1/1+logc*ab
Option 2 is correct.
Step-by-step explanation:
To find value of expression: \frac{1}{1+log_a\:bc}+\frac{1}{1+log_b\:ca}+\frac{1}{1+log_c\:ab}
1+log
a
bc
1
+
1+log
b
ca
1
+
1+log
c
ab
1
We use the following result,
log_c\:x=\frac{log\:x}{log\:c}\:and\:log\,xy=\log\,x+log\,ylog
c
x=
logc
logx
andlogxy=logx+logy
Consider,
\frac{1}{1+log_a\:bc}+\frac{1}{1+log_b\:ca}+\frac{1}{1+log_c\:ab}
1+log
a
bc
1
+
1+log
b
ca
1
+
1+log
c
ab
1
\implies\frac{1}{1+\frac{log\,bc}{log\,a}}+\frac{1}{1+\frac{log\,ca}{log\,b}}+\frac{1}{1+\frac{log\,ab}{log\,c}}⟹
1+
loga
logbc
1
+
1+
logb
logca
1
+
1+
logc
logab
1
\implies\frac{log\,a}{log\,a+log\,bc}+\frac{log\,b}{log\,b+log\,ca}+\frac{log\,c}{log\,c+log\,ab}⟹
loga+logbc
loga
+
logb+logca
logb
+
logc+logab
logc
\implies\frac{log\,a}{log\,a+log\,b+log\,c}+\frac{log\,b}{log\,b+log\,c+log\,a}+\frac{log\,c}{log\,c+log\,a+log\,b}⟹
loga+logb+logc
loga
+
logb+logc+loga
logb
+
logc+loga+logb
logc
\implies\frac{log\,a+log\,b+log\,c}{log\,b+log\,c+log\,a}⟹
logb+logc+loga
loga+logb+logc
\implies1⟹1
Therefore, Option 2 is correct