Main content

## Algebra 1

### Course: Algebra 1 > Unit 12

Lesson 2: Exponential expressions- Exponential expressions word problems (numerical)
- Exponential expressions word problems (numerical)
- Initial value & common ratio of exponential functions
- Exponential expressions word problems (algebraic)
- Exponential expressions word problems (algebraic)
- Interpreting exponential expression word problem
- Interpret exponential expressions word problems

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Initial value & common ratio of exponential functions

CCSS.Math:

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.

## Want to join the conversation?

- Is "growth rate" another way of saying common ratio or is it something different?(19 votes)
- You are correct. The growth rate is another way of calling it the common ratio.(16 votes)

- is another name of the common ratio the y-intercept?(13 votes)
- 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 in`x = 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.(22 votes)

- I don't get how the ratios between h(1) and h(0) , and h(2) with h(1) both equal 2(7 votes)
- 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 ∙`2`

⁰ = 1

h(1) = ¼ ∙`2`

¹ = 2

h(2) = ¼ ∙`2`

² = 4

and h(n) = ¼ ∙`2`

ⁿ

In each case the ∙`2`

is visible in the original formula and all of the formulas we set up for evaluating the function(17 votes)

- Is there a way to figure out...

5/180 = 6^n

...without just guessing and guessing?(7 votes)- 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(8 votes)

- 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.(6 votes)- 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(7 votes)

- Please help!

what is the common ratio?

2, -6, 18, -54, ...?(5 votes)- 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*.(6 votes)

- at5:08

isn't this the formula of Explicit geometric sequence ?(3 votes)- In a geometric sequence, the input value is multiplied. Here it is the power.(3 votes)

- Are exponential functions and geometric sequences basically the same thing?(2 votes)
- In a way yes, geometric sequences are a specific type of exponential function. Whether they're in recursive or explicit form they have a formula that you fill in the blanks for that's slightly different then other functions.(5 votes)

- can you just do a video on just graphing exponential functions please(4 votes)
- At4:10, how did
**2^n+1/ 2^n= 2**(3 votes)- When we divide with exponents, we subtract them.

For example: 2^5 / 2^4 = 2^(5-4) = 2^1 = 1.

Sal did essentially the same thing: 2^(n+1) / 2^n = 2^(n+1-n) = 2^1 = 2

Hope this helps.(4 votes)

## Video transcript

- [Voiceover] So let's
think about a function. I'll just give an example. Let's say, h of n is equal to one-fourth times two to the n. So, first of all, you might notice something
interesting here. We have the variable, the
input into our function. It's in the exponent. And a function like this is
called an exponential function. So this is an exponential. Ex-po-nen-tial. Exponential function, and that's because the variable, the input into our function, is sitting in its definition of what is the output of
that function going to be. The input is in the exponent. I could write another
exponential function. I could write, f of, let's say the input is a variable, t, is equal to is equal to five times times three to the t. Once again, this is an exponential function. Now there's a couple of
interesting things to think about in exponential function. In fact, we'll explore many of them, but I'll get a little
used to the terminology, so one thing that you might see is a notion of an initial value. In-i-tial Intitial value. And this is essentially
the value of the function when the input is zero. So, for in these cases, the initial value for the function, h, is going to be, h of zero. And when we evaluate that, that's going to be one-fourth times two to the zero. Well, two to the zero power, is just one. So it's equal to one-fourth. So the initial value,
at least in this case, it seems to just be that number that sits out here. We have the initial value times some number to this exponent. And we'll come up with
the name for this number. Well let's see if this was
true over here for, f of t. So, if we look at its intial value, f of zero is going to be five times three to the zero power and, the same thing again. Three to the zero is just one. Five times one is just five. So the initial value is once again, that. So if you have exponential functions of
this form, it makes sense. Your initial value, well if you put a zero
in for the exponent, then the number raised to the exponent is just going to be one, and you're just going to be left with that thing that you're
multiplying by that. Hopefully that makes sense, but since you're looking at it, hopefully it does make a little bit. Now, you might be saying, well what do we call this number? What do we call that number there? Or that number there? And that's called the common ratio. The common common ratio. And in my brain, we say well why is it
called a common ratio? Well, if you thought about
integer inputs into this, especially sequential
integer inputs into it, you would see a pattern. For example, h of, let me do this in that green color, h of zero is equal to, we already
established one-fourth. Now, what is h of one going to be equal to? It's going to be one-fourth times two to the first power. So it's going to be one-fourth times two. What is h of two going to be equal to? Well, it's going to be one-fourth times two squared, so it's going to be times two times two. Or, we could just view this as this is going to be two times h of one. And actually I should have done this when I wrote this one out, but this we can write as two times h of zero. So notice, if we were to
take the ratio between h of two and h of one, it would be two. If we were to take the ratio between h of one and h of zero, it would be two. That is the common ratio between successive whole number
inputs into our function. So, h of I could say h of n plus one over h of n is going to be equal to is going to be equal to actually I can work it out mathematically. One-fourth times two to the n plus one over one-fourth times two to the n. That cancels. Two to the n plus one,
divided by two to the n is just going to be equal to two. That is your common ratio. So for the function h. For the function f, our common ratio is three. If we were to go the other way around, if someone said, hey, I have some function whose initial value, so let's say, I have some function, I'll
do this in a new color, I have some function, g, and we know that its initial initial value is five. And someone were to say its common ratio its common ratio is six, what would this exponential
function look like? And they're telling you this
is an exponential function. Well, g of let's say x is the input, is going to be equal to our initial value, which is five. That's not a negative sign there, Our initial value is five. I'll write equals to make that clear. And then times our common ratio to the x power. So once again, initial value, right over
there, that's the five. And then our common ratio is the six, right over there. So hopefully that gets
you a little bit familiar with some of the parts of an exponential function, why they are called what they are called.