Animated Solution for Mathematics - Quadratic Equations: The number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$ is :

The equation $5+|2^x-1|=2^x(2^x-2)$ appears intimidating, but we can simplify the landscape by identifying the recurring term $2^x$.

Let us perform a substitution: let $t=2^x$. This transforms our equation into:

Before proceeding, we must establish our ground rules. Since $t=2^x$ and $2^x$ is always strictly positive for any real $x$, we must enforce the constraint $t>0$.

This is the most critical step; forgetting this constraint is a common pitfall in the exam hall.

We now face the modulus term $|t-1|$. The modulus changes its behavior based on the value of $t$, with the critical point being $t=1$.

If $t\ge 1$, the expression inside is non-negative, so $|t-1|=t-1$.

If $t<1$, the expression is negative, so $|t-1|=-(t-1)=1-t$. This creates two distinct paths for our journey.

Let us explore Case 1: $t\ge 1$. Our equation becomes:

Simplifying this, we get $t+4=t^2-2t$, which rearranges into the quadratic:

Factoring this, we find $(t-4)(t+1)=0$. This gives us two potential roots: $t=4$ and $t=-1$.

Now, we apply our filter: does $t=4$ satisfy $t\ge 1$? Yes. Does $t=-1$ satisfy $t\ge 1$? No.

Thus, $t=4$ is our first valid victory.

Now, consider Case 2: $0<t<1$. Our equation becomes:

This simplifies to $6-t=t^2-2t$, or:

Factoring gives $(t-3)(t+2)=0$. The potential roots are $t=3$ and $t=-2$.

Again, we apply our filter: does $t=3$ satisfy $0<t<1$? No. Does $t=-2$ satisfy $0<t<1$? No. Both are rejected.

We have navigated the labyrinth and found only one valid value for $t$, which is $t=4$.

Our journey is not yet complete; we must return to the world of $x$. Since $t=2^x$, we have:

Recognizing that $4=2^2$, we conclude that $x=2$.

There is exactly one real root. You have successfully conquered the modulus, the exponential, and the domain constraints.