Animated Solution for Mathematics - Differentiation: Let $f:\mathbb{R}\to \mathbb{R}$ satisfy $f(x+y)=2^{x}f(y)+4^{y}f(x),\forall x,y\in \mathbb{R}$. If $f(2)=3$, then $14\cdot \frac{f^{\prime }(4)}{f^{\prime }(2)}$ is equal to

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a functional equation that might look like a tangled knot of variables and exponents at first glance.

We are given $f(x+y)=2^{x}f(y)+4^{y}f(x)$ with the condition $f(2)=3$. Our mission is to find the value of $14\cdot \frac{f^{\prime }(4)}{f^{\prime }(2)}$.

It might seem intimidating, but every functional equation is just a puzzle waiting for the right key. Let us unlock it together.

Look closely at the left-hand side of our equation: $f(x+y)$. In the realm of real numbers, addition is commutative, meaning $x+y$ is exactly the same as $y+x$.

Because the left-hand side is symmetric, the right-hand side must also be symmetric. If we swap $x$ and $y$ in the original equation, the left side remains $f(y+x)$, which is just $f(x+y)$.

Therefore, the right-hand side must also remain unchanged. Let us write this down:

This symmetry is the key that breaks the lock. We have successfully created a bridge between the two sides of the equation.

Now, we need to isolate our variables. We want all the $f(x)$ terms on one side and all the $f(y)$ terms on the other. Let us rearrange our symmetric equation:

If we divide both sides by $(4^x-2^x)(4^y-2^y)$, we get a beautiful, clean separation:

Since this equality holds for all $x$ and $y$, both sides must be equal to a constant, which we will call $c$. This gives us the general form of our function: $f(x)=c(4^x-2^x)$.

We are halfway there! We have the general form, but we need the specific value of $c$. We are given the condition $f(2)=3$. Let us plug $x=2$ into our function:

Since $f(2)=3$, we have $12c=3$, which means $c=\frac{3}{12}=\frac{1}{4}$. Our hidden function is finally revealed:

Now, we enter the final phase. We need the derivatives $f^{\prime }(4)$ and $f^{\prime }(2)$. Recall that the derivative of $a^x$ is $a^x\ln a$. Differentiating our function, we get:

Since $\ln 4=\ln (2^2)=2\ln 2$, we can simplify this to:

Now, let us calculate the values at $x=2$ and $x=4$. For $f^{\prime }(2)$:

For $f^{\prime }(4)$:

Finally, we compute the requested ratio:

And there it is! The $\ln 2$ terms vanished, the numbers aligned, and we arrived at 248. You have successfully navigated the symmetry, the separation, and the calculus.