Animated Solution for Mathematics - Basic Mathematics: The sum of the roots of the equation $x+1-2{\log }_2(3+2^x)+2{\log }_4(10-2^{-x})=0$, is :

Conquering the Logarithmic Maze

Welcome, fellow traveler on the JEE journey. Today, we are going to dissect a problem that, at first glance, looks like a chaotic mess of bases and exponents.

You see an equation like $x+1-2{\log }_2(3+2^x)+2{\log }_4(10-2^{-x})=0$ and your instinct might be to panic. But I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask for elegance. Our job is to peel back that mask.

The Philosophy of Uniformity

Why do we struggle with this equation? It is because it speaks three different languages: linear algebra ($x+1$), base-2 logarithms, and base-4 logarithms. To solve it, we must force these terms to speak the same language.

We choose base 2 because it is the most fundamental unit here. First, let us tackle the linear term $x+1$. We know that any number $n$ can be expressed as ${\log }_a(a^n)$. Thus, $x+1$ becomes ${\log }_2(2^{x+1})$.

Now, look at the term $2{\log }_4(10-2^{-x})$. The base 4 is simply $2^2$. Using the base change property ${\log }_{a^k}b=\frac{1}{k}{\log }_{a}b$, we can pull that power out.

The $2$ in front and the $\frac{1}{2}$ from the base cancel out perfectly, leaving us with ${\log }_2(10-2^{-x})$. Suddenly, the equation is not a mess anymore; it is a symphony of base-2 logarithms.

The Logarithmic Dance

Now that we have ${\log }_2(2^{x+1})-{\log }_2(3+2^x{)}^2+{\log }_2(10-2^{-x})=0$, we can use the fundamental laws of logarithms. Remember, $\log A-\log B+\log C=\log \left(\frac{A\cdot C}{B}\right)$.

We combine everything into one single, powerful expression:

This is the moment of truth. If the logarithm of an expression is zero, the expression itself must be $2^0$, which is $1$. We have successfully stripped away the logarithmic layer, leaving us with a purely algebraic fraction:

The Algebraic Metamorphosis

This is where many students get stuck, staring at the powers of $2$. But look closely—we have $2^x$ everywhere. Let us perform a substitution: $t=2^x$.

This transforms our equation into:

Watch the numerator simplify. $2t\cdot 10$ is $20t$, and $2t\cdot \frac{1}{t}$ is just $2$. So, we have $20t-2=(3+t{)}^2$.

Expanding the right side gives us $9+t^2+6t$. Rearranging everything to one side, we arrive at the beautiful quadratic:

The Elegant Conclusion

We are almost there. We have a quadratic equation $t^2-14t+11=0$. Its roots are $t_1=2^{x_1}$ and $t_2=2^{x_2}$.

We do not need to solve for $t$ using the quadratic formula! That would be a waste of time. We need the sum of the roots $x_1+x_2$.

From Vieta's formulas, we know the product of the roots $t_1\cdot t_2=\frac{c}{a}=11$. Since $t_1=2^{x_1}$ and $t_2=2^{x_2}$, their product is $2^{x_1}\cdot 2^{x_2}=2^{x_1+x_2}$.

Therefore, $2^{x_1+x_2}=11$. Taking the logarithm base 2 on both sides, we get the final result:

See? The complexity vanished. We didn't fight the equation; we guided it to its natural conclusion. Keep this mindset, and no JEE problem will ever intimidate you again.