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)=2^(x^(2))*5^(x^(2))`
` or 5^(6-x-x^(2))=2^(x^(2)-x-2)`
`or (6-x-x^(2))log_(10)5 = (x^(2)-x-2)log_(10)2`(base 10)
` or (6-x-x^(2))[1-log_(10)2]=(x^(2)-x-2)log_(10)2`
` or 6 - x-x^(2) = ( log _(10) 2) [(x^(2) - x - 2) - x^(2) - x + 6]`
` or 6 - x - x^(2) = (log_(10)2)[4-2x]`
` or x^(2) + x - 6 = 2 (log_(10)2)(x-2)`
` or (x+3)(x-2) = (log_(10)4) (x-2)`
Therefore, either ` x = 2 or x + 3 = log_(10) 4`
`rArr x = log_(10) 4 - 3 = log_(10) (4/(1000))`
` rArr x =- log_(10) (250)`