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

Peeling Back the Layers

Mathematics, much like life, often presents us with problems that seem overwhelming at first glance. When you look at the equation ${\log }_7{\log }_5(\sqrt{x+5}+\sqrt{x})=0$, it is natural to feel a moment of hesitation.

It looks like a complex, nested structure. But here is the secret: treat it like an onion. We don't need to tackle the whole thing at once; we just need to peel it, one layer at a time.

The Outer Shell

Our goal is to isolate $x$. The outermost layer is the logarithm with base $7$.

We know that ${\log }_7(A)=0$. By the fundamental definition of a logarithm, ${\log }_b(A)=C$ is equivalent to $A=b^C$. Here, our base $b$ is $7$, our argument $A$ is the entire inner expression ${\log }_5(\sqrt{x+5}+\sqrt{x})$, and our exponent $C$ is $0$.

Therefore, we can write:

Since any non-zero number raised to the power of $0$ is $1$, our equation simplifies beautifully to:

The Inner Core

Now, we have peeled away the first layer. We are left with ${\log }_5(\sqrt{x+5}+\sqrt{x})=1$.

We apply the same logic again. Our base $b$ is $5$, our argument $A$ is $\sqrt{x+5}+\sqrt{x}$, and our exponent $C$ is $1$. This gives us:

Which is simply:

The Radical Dance

We have successfully reduced a complex logarithmic equation to a classic radical equation. Now, we must be strategic.

If we square both sides immediately, we will end up with a cross-product term $2\sqrt{x(x+5)}$ that keeps the radicals alive. Instead, let's isolate one radical by subtracting $\sqrt{x}$ from both sides:

Now, we square both sides. This is the moment of truth where the algebra clears the path for us:

Expanding the right side using the identity $(a-b{)}^2=a^2-2ab+b^2$, we get:

Look at that! The $x$ terms on both sides cancel out perfectly. We are left with a simple linear equation in terms of the radical:

Rearranging to solve for $\sqrt{x}$:

Finally, squaring both sides one last time gives us $x=4$.

In the world of JEE Advanced, we never just stop at the answer. We verify.

If we plug $x=4$ back into our original equation, we get $\sqrt{4+5}+\sqrt{4}=3+2=5$. Then, ${\log }_5(5)=1$, and ${\log }_7(1)=0$.

The logic holds, the math is consistent, and the solution is solid. You have successfully navigated the layers of this problem. Remember, no matter how complex the expression, stay calm, peel the layers, and trust the fundamental definitions.