The real solutions of the equation `2^(x+2). 5^(6-x)=10^x^2` is/are 1 (b) 2 (c) `-(log)_(10)(250)` (d) `(log)_(10)4-3`

Options

A. `1`
B. `2`
C. `-log_(10)(250)`
D. `log_(10)4 - 3`

Answer

B, C, D

Explanation

`2^(x + 2), 5^(6 - x) = 10^(x^(2)) = 5^(x^(2)) . 2^(x^(2))`
`rArr 5^(6 - x-x^(2)) = 2^(x^(2) - x - 2)`
take `log` these `10` on both sides,
`(6-x-x^(2))log_(10)5=(x^(2)-x-2)log_(10)2`
`(6-x-x^(2))(1-log_(10)2) = (x^(2)-x-2)log_(10)2`
`(6-x-x^(2)) = log_(10)2(x(2)-x-2)+(6-x-x^(2))`
`rArr (6-x-x^(2)) = (log_(10)2)(4-2x)`
`rArr x^(2) + x -6 = 2(log_(10)2)(4 - 2x)`
`rArr (x + 3)(x - 2) = (2log_(10)2)(x - 2)`
`rArr (x - 2)(x + 3 - log_(10)4) = 0`
`:. x = 2` or `x = log_(10)4 - 3`
`x = log_(10)4 - log_(10)1000`
`= log_(10)((4)/(1000)) = log_(10)((1)/(250))`
`x = -log_(10)(250)`