BdMO Online Forum

no. of real solution of the equation $4^x = x^2$

The given equation splits up into $2$ different equations.
$2^x=x$ for positive $x$. and $2^x=-x$ for negative $x$.
That means the equation we have to solve is this $2^x=|x|$
When $x$ is negative the left side is less than $1$ and tending to $0$.So no solution in this case.
If $x$ is positive the graph is a convex function starting from $0$.where the line $y=x$ is below the graph.So they never intersect.

zadid xcalibured wrote: When $x$ is negative the left side is less than $1$ and tending to $0$.So no solution in this case.

This statement is not quite right, there is one intersection, as RHS goes from $0$ to $\infty$ and LHS goes from $1$ to $0$ for $x= 0 \text{ to } -\infty $, and as both are strictly increasing/decreasing functions.

How could i do this kind of shitty thing?????????? I don't know who gave me the right to turn a curve into a straight line. The first mistake is where i eliminated the square. But the problem is pretty easy.

zadid xcalibured wrote:I don't know who gave me the right to turn a curve into a straight line. The first mistake is where i eliminated the square.

I don't think that was a mistake, actually, it was a nice idea to eliminate the square.