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Step-by-Step Solution
Step 1: Write down the given functional equation
We have
f(x + y) = 2^x \, f(y) + 4^y \, f(x), \quad \forall\, x,y \in \mathbb{R}.
Step 2: Derive a relationship for f(x)
By swapping x and y , we also get
f(y + x) = 2^y \, f(x) + 4^x \, f(y).
Since f(x + y) = f(y + x), it follows that
2^x \, f(y) + 4^y \, f(x) \;=\; 2^y \, f(x) + 4^x \, f(y).
Rearranging gives
(4^y \;-\; 2^y)\, f(x) \;=\; (4^x \;-\; 2^x)\, f(y).
This implies
\frac{f(x)}{4^x - 2^x} \;=\; \frac{f(y)}{4^y - 2^y} \;=\; k,
where k is a constant. Hence,
f(x) = k\,(\,4^x - 2^x\,).
Step 3: Determine the constant k using f(2) = 3
Plug in x = 2 :
f(2) = k \,\bigl(4^2 - 2^2\bigr) = k\,(16 - 4) = 12\,k.
Given f(2)=3 , so 12\,k = 3 \implies k = \frac{1}{4}.
Thus,
f(x) = \frac{4^x - 2^x}{4}.
Step 4: Differentiate f(x)
We differentiate term by term:
f'(x) = \frac{1}{4} \bigl(4^x \ln(4) - 2^x \ln(2)\bigr).
Recall that \ln(4) = 2\,\ln(2).
Step 5: Compute f'(4) and f'(2)
1) For x=4 :
f'(4)
= \frac{1}{4}\bigl(4^4 \cdot \ln(4) - 2^4 \cdot \ln(2)\bigr)
= \frac{1}{4}\bigl(256 \cdot 2\,\ln(2) - 16 \,\ln(2)\bigr).
f'(4)
= \frac{1}{4}\bigl(512\,\ln(2) - 16\,\ln(2)\bigr)
= \frac{496\,\ln(2)}{4}
= 124\,\ln(2).
2) For x=2 :
f'(2)
= \frac{1}{4}\bigl(4^2 \cdot \ln(4) - 2^2 \cdot \ln(2)\bigr)
= \frac{1}{4}\bigl(16 \cdot 2\,\ln(2) - 4\,\ln(2)\bigr).
f'(2)
= \frac{1}{4}\bigl(32\,\ln(2) - 4\,\ln(2)\bigr)
= \frac{28\,\ln(2)}{4}
= 7\,\ln(2).
Step 6: Find the ratio \,\frac{f'(4)}{f'(2)}\,
\frac{f'(4)}{f'(2)} = \frac{124\,\ln(2)}{7\,\ln(2)} = \frac{124}{7}.
Step 7: Multiply the ratio by 14
We need 14 \times \frac{f'(4)}{f'(2)} , so
14 \times \frac{124}{7} = 14 \times \frac{124}{7} = 2 \times 124 = 248.
Therefore, the value of
14 \,\frac{f'(4)}{f'(2)}
is
\boxed{248}.
