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# Initial value & common ratio of exponential functions

In an exponential function of the form f(x) = a*b^x, the initial value is usually taken to be the value of f(0), or "a". The common ratio refers to the rate of change in an exponential function. In the form given above, the common ratio is "b". For example, in the function f(x) = 2*3^x, the initial value is 2 and the common ratio is 3.

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• Is "growth rate" another way of saying common ratio or is it something different?
• You are correct. The growth rate is another way of calling it the common ratio.
• is another name of the common ratio the y-intercept?
• No, the y-intercept is the value when x is equal to 0. For an exponential function the y-intercept is the "initial value" not the common ratio.
Consider a standard exponential function of the form y(x) = a•rˣ, if you put inx = 0 you get: y(0) = a•rˣ = a•r⁰ = a•1 = a, so the y-intercept is a, which is called the initial value, not r, which is called the common ratio.
• I don't get how the ratios between h(1) and h(0) , and h(2) with h(1) both equal 2
• Well, if h(x) = ¼ ∙2ⁿ,
then as Sal says,
h(0) = ¼ ∙2⁰ which means that h(0) = ¼ ∙1 because 2⁰ = 1
and h(1) = ¼ ∙2¹ which means that h(1) = ¼ ∙2¹ because 2¹ = 2
and h(2) = ¼ ∙2² which means that h(2) = ¼ ∙2 ∙2 → because 2² = 2 ∙2 = 4

Lining them up without all the explanations,
h(0) = ¼ ∙2⁰ = 1
h(1) = ¼ ∙2¹ = 2
h(2) = ¼ ∙2² = 4

To find the ratios, we divide each result by the previous result, so h(2) ÷ h(1) = 4 ÷ 2 = 2
and, h(1) ÷ h(0) = 2 ÷ 1 = 2

Another way to look at this is
h(0) = ¼ ∙2⁰ = 1
h(1) = ¼ ∙2¹ = 2
h(2) = ¼ ∙2² = 4

and h(n) = ¼ ∙2
In each case the ∙2 is visible in the original formula and all of the formulas we set up for evaluating the function
• Is there a way to figure out...
5/180 = 6^n
...without just guessing and guessing?
• 6^n = 1/36
log_6(6^n) = log_6(1/36)
n = log_6(1/36) = log(1/36)/log(6) ("log" = log_10 (= "log") or log_e (= "ln") on a calculator).
n = -2

Or you could reason:

6^n = 1/36
Since 6^2 is 36, then 1/36 is 1/(6^2) = 6^(-2)
Therefore log_6(1/36) is -2
• Im trying some problems but in the hints, they dont show a step and it's completely shutting me down on every problem.

(1/3)^n = 27 --> (1/3)^2 = 3^3 --> (1/3)^n = (1/3)^3 --> n = -3

please help me. Im getting stuck on all the problems at this step. Everything prior is OK, its just here I get stuck and dont know the steps/process.
• I assume you are trying to find "n".
When we look at: (1/3)^n = 27, we need to think about 2 things.
1) Does "n" need to be positive or negative?
2) What exponent will change 3 into 27?

1) Does "n" need to be positive or negative? I try to do this first. Notice, you are starting with 1/3. If we apply positive exponents, we always get a fraction with a numerator of 1: (1/3)^2 = 1/9; (1/3)^3 = 1/27; etc. But, we need 27 in the numerator. This tells us the exponent needs to be negative. We need to flip the fraction to get the 27 on top.

2) What exponent will change 3 into 27? After step one, this is usually the easy part.
3^3 = 27.

Putting the 2 steps together, we need n = -3
what is the common ratio?
2, -6, 18, -54, ...?
• To find the common ratio, we need to find a relationship that gets us from one number to the next. After looking at the first four numbers, we notice that they alternate from positive to negative. That means the common ratio is negative. And if we divide two consecutive numbers, say, 2 and -6, we find the number -3. It's negative, so to check, we multiply all the numbers by -3, and see if the answer is the next term:
2 * -3 = -6
-6 * -3 = 18
18 * -3 = -54
It's all correct, so the common ratio is -3.
• These are geometrics sequences
(1 vote)
• Good observations. They are similar, but not exactly the same.
The equation for a geometric sequence restricts the domain (input values) to counting numbers. A sequence of a(n) would have values of n = 1, 2, 3, 4, etc. because "n" represents the term in the sequence. If you graph the sequence in a coordinate plane, it would be a set of points with gaps in between.

An exponential function f(x) would have a domain (set of input values) which is any real number. When you graph an exponential function, it is a continuous graph (no gaps).

Hope this helps.
• Are exponential functions and geometric sequences basically the same thing?