Question

" Find the mid point of AB, where A(log2^8,log2^25)
and B(log10^10,log(10^100)​

Answer 1

Answer:

$\\\tt ( 2 , log_{4}(100) )$

Step-by-step explanation:

Coordinates of A:

$\\\tt (log_{2}(8), log_{2}(25) ) \\ = \ \ \tt(3, log_{2}(25) )$

Coordinates of B:

$\\\tt (log_{10}(10), log_{10}(100) ) \\\tt = (1, 2 )$

Mid point is given by:

$\\\tt ( \frac{x_{1}+x_{2} }{2} , \frac{y_{1}+y_{2} }{2})$

Using that here,

$\\\tt ( \frac{3+1 }{2} , \frac{ log_{2}(25) + 2}{2 }) \: \\\tt ( 2 , \frac{ log_{2}(25) + 2}{2 })$

You can and the answer here or go on collapse y into a single log as shown

$\\\tt ( 2 , \frac{ log_{2}(25) + log_{2}(4) }{ log_{2}(4) }) \\\tt = ( 2 , \frac { log_{2}(100) }{ log_{2}(4) }) \\\tt = ( 2 , log_{4}(100) )$