The roots of the equation 4^x - 3 cdot 2^{x+2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given equation is: $4^x - 3 \cdot 2^{x+2} + 32 = 0$Since $4^x = (2^2)^x = (2^x)^2$ and $2^{x+2} = 2^x \cdot 2^2 = 4 \cdot 2^x$,the equation become

Vedclass · Accepted Answer

$2$ and $3$

Vedclass · Answer

(D) Given equation is: $4^x - 3 \cdot 2^{x+2} + 32 = 0$Since $4^x = (2^2)^x = (2^x)^2$ and $2^{x+2} = 2^x \cdot 2^2 = 4 \cdot 2^x$,the equation becomes:$(2^x)^2 - 3 \cdot (4 \cdot 2^x) + 32 = 0$$(2^x)^2 - 12 \cdot 2^x + 32 = 0$Let $2^x = y$. Then the equation is $y^2 - 12y + 32 = 0$.Factoring the quadratic equation: $(y - 8)(y - 4) = 0$.So,$y = 8$ or $y = 4$.Substituting $y = 2^x$ back:$2^x = 8 \Rightarrow 2^x = 2^3 \Rightarrow x = 3$$2^x = 4 \Rightarrow 2^x = 2^2 \Rightarrow x = 2$Thus,the roots are $2$ and $3$.

The roots of the equation $4^x - 3 \cdot 2^{x+2} + 32 = 0$ are

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $6x^2 + 77x + 121 = 0$
$II.$ $y^2 + 9y - 22 = 0$

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{3} - 468 = 1729$
$II.$ $y^{2} - 1733 + 1564 = 0$

The roots of the given equation $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$ are

Which of the following equations has equal roots?

If the two roots of the equation $(a - 1)(x^4 + x^2 + 1) + (a + 1)(x^2 + x + 1)^2 = 0$ are real and distinct,then the set of all values of $a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module