Question

Which exponential function has a faster rate f(x)=5^x or g(x)=(1/3)^x. How do you know?

Answer 1

f(x)

Explanation:

This question isn't to difficult, since all you need to know is:
`((a)/(b))^x=(a^x)/(b^x)`. If the you're not quite sure why this is true, I'll try to express this in another way.
`((a)/(b))^x=(a)/(b)*(a)/(b)*(a)/(b)...\text{ x amount of times}`, and as you may know when multiplying fractions, you simply multiply the numerators by the numerators and the denominators by the denominators, this means that you can simplify it to:
`\frac{a*a*a...\text{ x amount of times}}{b*b*b...\text{ x amount of times}}` this is literally what an exponent is which is why you can distribute the exponent as such:
`((a)/(b))^x=(a^x)/(b^x)`. The next thing you need to know is that:
`1^x=1` for any real number. This is you can express this as:
`1*1*1...\text{ x amount of times} = 1`, since 1 times 1 times 1, will always remain 1. You may think this will not apply for rational exponents, but it still does since:
`1^{(a)/(b)}=\sqrt[b]{1}^a`, and no matter what b is, it's going to be 1. The square root of 1, is 1..., the cube root of 1 is 1.... This is because the very definition of a radical is what number times times it self index amount of times is equal to the number under the radical, and no matter how many times you multiply 1, you're going to get 1. If that's a bit confusing what I'm saying is:
`\sqrt[n]{b}\implies\text{?}^n=b`.

Anyways hopefully you understand that, if not I'm sorry if I went a bit overboard trying to explain it. Anyways all you need to really know is that:
`((1)/(3))^x=(1^x)/(3^x)=(1)/(3^x)`. So this means as x increases, all you're really doing is increasing the denominator which is decreasing the number. The f function has an integer as a base, so the function will overall be increasing. This is not true for g(x) since as x increases the y-value decreases. So this means that f(x) has a faster rate. This is because the g(x) has a horizontal asymptote, since as x approaches infinity, g(x) approaches 0, since the denominator is increasing towards infinity. So the rate of change will be slowing down as x increases, since there is only so much the g(x) can change, since it has a horizontal asymptote. This is not true for f(x) as it has no horizontal asymptote and just goes towards infinity. So the f(x) has a faster rate.