If A(log28, log525) and B(log1010, log10100), then the midpoint of AB is
Answers
Answer:
SOLUTION
TO DETERMINE
The mid point of AB, where
\sf{A ( log_{2}8, log_{5}25 ) \: \: B( log_{10}10, log_{10}100 )}A(log
2
8,log
5
25)B(log
10
10,log
10
100)
FORMULA TO BE IMPLEMENTED
For the given two points
\sf{A( x_1 , y_1) \: \: and \: \: B( x_2 , y_2)}A(x
1
,y
1
)andB(x
2
,y
2
)
The midpoint of the line AB is
\displaystyle \sf{ \bigg( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} \bigg)}(
2
x
1
+x
2
,
2
y
1
+y
2
)
EVALUATION
Here the given points are
\sf{A ( log_{2}8, log_{5}25 ) \: \: B( log_{10}10, log_{10}100 )}A(log
2
8,log
5
25)B(log
10
10,log
10
100)
Now
\sf{ log_{2}(8) = log_{2}( {2}^{3} ) = 3 \: log_{2}(2) = 3 }log
2
(8)=log
2
(2
3
)=3log
2
(2)=3
\sf{ log_{5}(25) = log_{5}( {5}^{2} ) = 2 \: log_{5}(5) = 2 }log
5
(25)=log
5
(5
2
)=2log
5
(5)=2
\sf{ log_{10}(10) = 1 }log
10
(10)=1
\sf{ log_{10}(100) = log_{10}( {10}^{2} ) = 2 \: log_{10}(10) = 2 }log
10
(100)=log
10
(10
2
)=2log
10
(10)=2
So the points are simplified to
A (3,2) , B (1,2)
Hence the required midpoint is
\displaystyle \sf{ = \bigg( \frac{3 + 1}{2} , \frac{2 + 2}{2} \bigg)}=(
2
3+1
,
2
2+2
)
\displaystyle \sf{ = \bigg( \frac{4}{2} , \frac{4}{2} \bigg)}=(
2
4
,
2
4
)
\displaystyle \sf{ = (2 , 2)}=(2,2)
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