Here's a screenshot of the problem:
Because the $x$-value is an exponent, then this must be an exponential function. By definition, an exponential function is where the independent variable (the $x$-value) is the exponent.
To write the function in the form $K(x) = ab^x$, I first converted the radical into its exponent form and moved the denominator over to the numerator to get
$K(x) = (3^x)(3^{-\frac{1}{2}})(6^{-x})$
I then simplified it further, following the exponent rules of $a^ma^n = a^{m + n}$
$K(x) = (3^{x - \frac{1}{2}})(6^{-x})$
However, as shown, my answers were incorrect. What did I do wrong?
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$\begingroup$Because the x-value is an exponent, then this must be an exponential function. By definition, an exponential function is where the independent variable (the x-value) is the exponent.
While you're not wrong in this respect, this only really holds if the function is in its simplest form. Consider, for instance,
$$f(x) = \frac{3^x}{3^x}$$
Sure, the variable $x$ is only in exponents here. However, if you simplify, you realize immediately $f(x) = 1$, which is obviously not exponential.
So we have to be very careful about how we define things as a result of things like these. We define $f$ to be an exponential function if there exist constants $a,b$ such that $f(x) = ab^x$. To show a function is exponential, then, you need to somehow manipulate it into that form and determine the constants $a,b$ necessary.
To reiterate: this is the real definition of an exponential function. (Well, to an extent; there are modifications to the definition you can make, but this is the most relevant one for your case.) The notion of "$x$ appears in the exponent" is an intuition, but clearly doesn't always hold.
So, with this mind, we need to figure out what constants $a,b$ you need for
$$f(x) = \frac{3^x}{\sqrt 3 \cdot 6^x}$$
Well, first note that, through exponent rules,
$$f(x) = \frac{1}{\sqrt{3}} \cdot \frac{3^x}{6^x} = \frac{1}{\sqrt 3} \cdot \left( \frac 3 6 \right)^x = \frac{1}{\sqrt 3} \cdot \left( \frac 1 2 \right)^x$$
With this manipulation, it's clear what your constants $a,b$ are, and thus you can conclude that $f$ is exponential.
Again, be very careful about what $a,b$ can be: this is where you went wrong. They have to be constants, not functions themselves. $3^{x-0.5}$ is not a constant. Basically, you cannot have your constants depend on $x$.
$\endgroup$ 2 $\begingroup$We have $K(x) = \frac{3^x}{\sqrt{3}\cdot6^x} = (\frac{3}{6})^x\cdot\frac{1}{\sqrt{3}} = (\frac{1}{2})^x \cdot \frac{1}{\sqrt{3}}$, so $a = \frac{1}{\sqrt{3}}$ and $b = \frac{1}{2}$.
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