NIMCET 2016 — Mathematics PYQ

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Question

NIMCET 2016 — Mathematics PYQ

NIMCET | Mathematics | 2016

Sum of the roots of the equation $4^{x} - 3 (2^{x + 3}) + 128 = 0$ is

Choose the correct answer:

Explanation

Concept:
Base Rule
If b raised to the xth power is equal to b raised to the yth power, that implies that x = y."
$b^{x} = b^{y} \Rightarrow x = y$
Calculations:
Given equation is $4^{x} - 3 (2^{x + 3}) + 128 = 0$
$\Rightarrow (2^{2})^{x} - 3 (2^{x} \cdot 2^{3}) + 128 = 0$
$\Rightarrow (2^{x})^{2} - 24 (2^{x}) + 128 = 0$
$\Rightarrow (2^{x})^{2} - 16 (2^{x}) - 8 (2^{x}) + 128 = 0$
$\Rightarrow (2^{x} - 16) (2^{x} - 8) = 0$
$\Rightarrow 2^{x} = 16$ or $2^{x} = 8$
$\Rightarrow 2^{x} = 2^{4}$ or $2^{x} = 2^{3}$
$\Rightarrow x = 4$ or $x = 3$
The roots of the equation $4^{x} - 3 (2^{x + 3}) + 128 = 0$ are 4 and 3
Its Sum = 4 + 3 = 7