Suppose f(x+y)=f(x)⋅f(y), f(x)=1+xg(x) where x→0 Ltg(x)=1 then the value of log f(8) is ___

Solution:

$\begin{array}{l} f (x + y) = f (x) f (y) \\ f (0) = f (0) f (0) \\ \Rightarrow f (0) = 0 or f (0) = 1 \end{array}$

And $f (x + 0) = f (x) f (0)$

If $f (0) = 0, then f (x) = f (x + 0) = f (x) 0 = 0$

$f (x) = 0 \forall x$

$\underset{x \to 0}{Lt} f (x) = 1 + \underset{x \to 0}{Lt} x \cdot g (x) = 1 + 0 = 1 contradicts f (x) = 0$

$∴ f (0) = 1$

$\begin{array}{l} \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{f (x) f (h) - f (x)}{h} \\ = \lim_{h \to 0} f (x) \frac{f (h) - 1}{h} \\ f' (x) = f (x) \lim_{h \to 0} \frac{f (h) - 1}{h} = f (x) \lim_{h \to 0} \frac{1 + hg (h) - 1}{h} = f (x) \lim_{h \to 0} \frac{hg (h)}{h} = f (x) \lim_{h \to 0} g (h) = f (x) \end{array}$

$\begin{matrix} f^{1} (x) = f (x) \\ \Rightarrow f (x) = c . e^{x} \Rightarrow f (0) = c = 1 \\ f (x) = e^{x} \\ Log f (x) = x \\ Log f (8) = 8 \end{matrix}$

Suppose $f (x + y) = f (x) \cdot f (y), f (x) = 1 + xg (x)$ where $_{x \to 0}^{L t} g (x) = 1$ then the value of log f(8) is ___

Solution:

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