The number of solutions of the equation 2^x = | vedclass

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

Question

(C) To find the number of solutions for the equation $2^x = x^2$, we can analyze the intersection points of the functions $f(x) = 2^x$ and $g(x) = x^2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) To find the number of solutions for the equation $2^x = x^2$, we can analyze the intersection points of the functions $f(x) = 2^x$ and $g(x) = x^2$.$1$. For $x $2$. For $x > 0$: We check specific integer values:At $x = 2$, $2^2 = 4$ and $2^2 = 4$. So, $x = 2$ is a solution.At $x = 4$, $2^4 = 16$ and $4^2 = 16$. So, $x = 4$ is a solution.$3$. Between $x = 2$ and $x = 4$, the function $x^2$ is greater than $2^x$ (e.g., at $x = 3$, $2^3 = 8$ and $3^2 = 9$).For $x > 4$, the exponential function $2^x$ grows much faster than the quadratic function $x^2$, so there are no more solutions for $x > 4$.Thus, the solutions are $x = -0.766$ (approx), $x = 2$, and $x = 4$. There are a total of $3$ solutions.

The number of solutions of the equation $2^x = x^2$ is

Similar Questions

If the roots of the equation $ax^2 + bx + c = 0$ are reciprocal to each other,then:

If the equations $2x^2 + 3x + 5\lambda = 0$ and $x^2 + 2x + 3\lambda = 0$ have a common root,then $\lambda = $

Solve the given two equations and select the correct answer from the given options.
$I.$ $\quad x = \sqrt[3]{2197}$
$II.$ $\quad y^2 = 169$

When $x+\frac{1}{x}=3$,find $x^{2}+\frac{1}{x^{2}}$.

Let two numbers have an arithmetic mean of $9$ and a geometric mean of $4$. Then these numbers are the roots of the quadratic equation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module