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The real solutions of the equation `2^(x+2). 5^(6-x)=10^x^2`is/are1 (b)2 (c) `-(log)_(10)(250)`(d) `(log)_(10)4-3`A. `1`B. `2`C. `-log_(10)(250)`D. `log_(10)4 - 3` |
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Answer» Correct Answer - B::C::D `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)` |
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