Animated Solution for Mathematics - Quadratic Equations: The sum of all the solutions of the equation $(8{)}^{2x}-16\cdot (8{)}^x+48=0$ is :

Welcome, future engineer. Today, we are going to dismantle a problem that often trips up students in the heat of a JEE Advanced exam. It looks like a terrifying exponential beast, but beneath that exterior lies a simple, elegant quadratic structure waiting to be revealed.

Look at the equation: $(8{)}^{2x}-16\cdot (8{)}^x+48=0$.

When you see an equation like this, your first instinct might be panic. But take a breath and look at the terms. We have $(8{)}^x$ and $(8{)}^{2x}$.

Recall the fundamental law of exponents: $(a^m{)}^n=a^{mn}$. This means that $(8{)}^{2x}$ is exactly the same as $((8{)}^x{)}^2$.

Suddenly, the equation transforms. It is no longer just an exponential mess; it is a quadratic equation in disguise.

Let us introduce a dummy variable, $t$, where $t=(8{)}^x$. By substituting this into our equation, we get:

See how the fear factor just vanished? We have moved from the world of exponents into the familiar, comfortable territory of algebra. This is the JEE mindset: identifying the hidden structure within the complexity.

Now, we need to solve for $t$. We are looking for two numbers that multiply to $48$ and add up to $-16$. Since the product is positive and the sum is negative, both numbers must be negative.

Think about the factors of $48$: $1\times 48$, $2\times 24$, $3\times 16$, $4\times 12$.

Ah, $4$ and $12$! Since $-4+(-12)=-16$ and $(-4)\cdot (-12)=48$, our factors are clear. We can write the equation as:

This gives us two solutions for $t$: $t=4$ and $t=12$. But remember, we are not solving for $t$; we are solving for $x$.

Now, we must perform the 'back-substitution.' We set our original $t$ back to $(8{)}^x$. This gives us two distinct cases:

1. $(8{)}^x=4$ 2. $(8{)}^x=12$

How do we extract $x$ from the exponent? We use the logarithm, the inverse operation of exponentiation. By definition, if $a^y=x$, then $y={\log }_a(x)$.

Applying this to our cases: For the first case: $x_1={\log }_8(4)$ For the second case: $x_2={\log }_8(12)$

The problem asks for the sum of all solutions. Let $S=x_1+x_2$. Therefore:

We are almost there. Recall the product rule for logarithms: ${\log }_b(m)+{\log }_b(n)={\log }_b(mn)$. Applying this to our sum:

We know that ${\log }_8(8)=1$. Can we extract an $8$ from $48$? Absolutely, since $48=8\cdot 6$.

Using the product rule in reverse, we split this into:

Since ${\log }_8(8)=1$, we arrive at our final, beautiful result:

By staying calm, identifying the structure, and using the right tools, even the most intimidating problems yield to your logic. Keep practicing this mindset, and you will master the JEE.