Answer
Both are exponential.
$f(x)=2.7\cdot 3^{x},\displaystyle \quad g(x)=0.75\cdot(\frac{1}{2})^{x}$
Work Step by Step
If a function x is an exponential function, then it has form
$f(x)=a\cdot b^{x}$
When we observe an increase of x by 1 unit, $f(x+1)=ab^{x+1}.$
what follows is that the ratio $\displaystyle \frac{f(x+1)}{f(x)}=\frac{ab^{x+1}}{ab^{x}}=b.$
(the ratio is constant when arguments increase by 1).
Thus, if the x increases by unit, the function value is multiplied with b.
We use this to recognize whether a table of function values represent an exponential function or not.
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The x-values increase by 1 unit.
For the values of f(x) we observe $ \displaystyle \frac{f(x+1)}{f(x)}$:
$\displaystyle \frac{0.9}{0.3}=3,\quad\frac{2.7}{0.9}=3,\quad\frac{8.1}{2.7}=3,\quad\frac{24.3}{8.1}=3,\quad$
so f is exponential, $f(x)=a\cdot 3^{x}$
Since $f(0)=2.7$, then $a=2.7$
Thus, $f(x)=2.7\cdot 3^{x}$
For the values of g(x),
$\displaystyle \frac{1.5}{3}=\frac{1}{2},\quad\frac{0.75}{1.5}=\frac{1}{2},\quad\frac{0.375}{0.75}=\frac{1}{2},\quad\frac{0.1875}{0.375}=\frac{1}{2}$
so it is exponential, $g(x)=a\displaystyle \cdot(\frac{1}{2})^{x}.$
Since $g(0)=0.75$, then $a=0.75$
Thus, $g(x)=0.75\displaystyle \cdot(\frac{1}{2})^{x},\qquad ($which can also be written as $\displaystyle \frac{3}{4}\cdot 2^{-x})$