Animated Solution for Mathematics - Basic Mathematics: The solution of equation ${\log }_7{\log }_5(\sqrt{x+5}+\sqrt{x})=0$ is .........

Solving Nested Logarithms

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that looks intimidating at first glance but reveals itself to be a beautiful exercise in logical sequencing.

We are looking at the equation:

At first, this looks like a complex, multi-layered structure. Visualize this as an onion; to reach the center—the value of $x$—we simply need to peel the layers one by one.

Peeling the Layers

The beauty of logarithms lies in their definition. We know that ${\log }_{a}b=c$ is equivalent to $b=a^c$. This is our master key.

Our outermost layer is the ${\log }_7$. To remove it, we apply the definition: the entire expression inside the ${\log }_7$ must equal $7^0$.

Since any non-zero number raised to the power of $0$ is $1$, we get:

Just like that, the first layer is gone! We are left with a much friendlier equation. Now, we apply the same logic to the ${\log }_5$. The base $5$ moves to the right side, taking $1$ as its exponent:

We have successfully peeled the onion. We are no longer dealing with logarithms; we are now in the realm of radical algebra.

The Radical Dance

Now, we face the equation $\sqrt{x+5}+\sqrt{x}=5$. Many students, in their haste, might try to square both sides immediately. I urge you to pause.

If you square the left side as it is, you will end up with a cross-term $2\sqrt{x(x+5)}$, which keeps the radicals alive and complicates the process. Instead, let us use the isolation strategy. Move one radical to the other side:

Now, when we square both sides, the magic happens. The left side becomes $x+5$. The right side, using the identity $(a-b{)}^2=a^2-2ab+b^2$, becomes:

Setting them equal, we get:

Look closely at this equation. The $x$ terms on both sides cancel out perfectly! We are left with:

Rearranging this, we find $10\sqrt{x}=20$, which simplifies beautifully to $\sqrt{x}=2$. Squaring one last time, we arrive at our destination: $x=4$.

The Final Verification

In the world of JEE, we never leave a problem without verifying our work. Let us plug $x=4$ back into our original equation.

The expression becomes $\sqrt{4+5}+\sqrt{4}=\sqrt{9}+2=3+2=5$. Then, ${\log }_5(5)=1$, and finally, ${\log }_7(1)=0$.

The logic holds, the math is sound, and the answer is confirmed. You have successfully navigated the nested layers and emerged victorious. Keep this mindset—peel the layers, isolate the complexity, and trust the process.