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Analyzing graphs of exponential functions: negative initial value

Given the graph of an exponential function with a negative initial value, Sal finds the formula of the function and solves an equation.

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  • primosaur ultimate style avatar for user Bill Zemon
    Is there an algebraic way of determining the unknown value of an exponent? For example, 5^x = 625, how can we algebraically determine x? In the video, Sal wrote the values of 5^1, 5^2, 5^3, and 5^4 from memory. But how can the value of the exponent be calculated?
    (25 votes)
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  • male robot johnny style avatar for user Mukhtar Otarbayev
    I couldn’t understand why did 1/5^x=1/625 become 5^x=625? Please, help.
    (12 votes)
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  • aqualine ultimate style avatar for user RedBlackandBlue
    At what do you think the siren is?
    (6 votes)
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  • orange juice squid orange style avatar for user Jaewoo
    Why do we derive the common ration by f(0) or f(1).. Why not f(0), f(1.5)?
    (5 votes)
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    • aqualine tree style avatar for user Judith Gibson
      The common ratio is found by dividing one term by the term before it.
      Using the formula, that means that we would divide one term ( ar^x) by the term before it ( ar^(x-1) ). to give us an answer of r.
      If you used f(1.5) when wanting to find the common ratio,
      you would start with the term ar^1.5 and
      you would need to divide by ar^.5 in order to still arrive at the answer of r.
      To answer your question, it makes things much simpler to stick with whole numbers!
      (7 votes)
  • mr pink orange style avatar for user Maan Younus
    How can we algebraically solve a problem like this: 5(to the x power) = 625
    what is the mathematical way to find x ?
    (4 votes)
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    • female robot grace style avatar for user Benjamin Clements
      logarithms are the inverse of exponents in the same way that division is the inverse of multiplication, given x^y, log_x x^y = y aka log base x of x to the power of y equals y, or x^log_x(y) = y aka x to the power of the log base x of y is y eg. 10^(log_10 10^3) = 10^3.

      log_5 (5^x) = x
      since we're saying that 5^x = 625 then
      log_5 (625) = x

      if log_5 (5^x) != log_5 (625) then 5^x != 625

      A useful exponent/log/power rule is that log_n x = log_10 x / log_10 n, this is particularly useful since many calculators only provide log_10 not log_<whatever you want> (if it's labeled log then it's log_10 aka log base 10, ln is the "natural log" or log_e, in general it works for any log base eg. log_y x / log_y n = log_n x, so log_5 625 = log_10 / log 5 = ln 625 / ln 5 = ...)

      a calculator will tell you that log_5 625 is 4.

      Note that logarithms are not limited to whole numbers, you can take the log of 142 and get 2.15... since 10 to the power of 2.15... is 142.
      (5 votes)
  • starky seed style avatar for user Annabelle Sawyer
    Is this graph growth or decay or neither? If a is negative and b is between 0 and 1, the graph is increasing, but hits the asymptote at zero. Does "a" have to be always positive to be classified as growth and decay ? Is there such a thing as negative growth and decay?
    (3 votes)
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  • starky ultimate style avatar for user Alessandro V. Santoro
    Let's see if someone can help me, I have a tricky one. I have understood that the base of an exponential function can't be negative, since with fractional exponential values, you'd end up with complex numbers: f(x) = -3^x and you evaluate x = 1/2, you get f(1/2) = -3^(1/2), which leads to complex numbers

    Now, then why when I tried f(x) = -3x in DESMOS graphing calculator I got a graph? How can this graph exist?

    P.S. Sorry if there was a better lecture for where to post this question, this is the best I found.
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      You made a common error in the way you entered the function.
      Remember -3^2 and (-3)^2 are different.
      -3^2 = -(3*3) = -9
      (-3)^2 = (-3)(-3)

      The base of the exponent in f(x)=-3^x is "3", not "-3". And the "-" in front takes the standard exponential function and reflects it across the x-axis. This is why you got a graph.

      To have a base of -3, you need to enter the function as: f(x)=(-3)^x

      Hope this helps.
      (2 votes)
  • male robot hal style avatar for user {OG}SpeedyPotato644🥔💨💨💨
    can this kind of graph curve into a parabola?
    (1 vote)
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  • duskpin ultimate style avatar for user HavicDoesMath
    In other words, when Sal was solving 5^x = 625, he was trying to solve the logarithmic function log-sub-5, 625? Just making sure I understand. Thanks!
    (2 votes)
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  • mr pants teal style avatar for user Qasim Hashmi
    What's the difference between the words "successive" and "consecutive" ?
    (1 vote)
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    • mr pink green style avatar for user David Severin
      If there is a difference, it must be pretty subtle. When talking about numbers, maybe the difference is that consecutive may be dealing more with whole numbers, 34-35-36-37 whereas successive just means in order, so if you have a table of values of x divided in tenths, successive would be 0-.1-.2-.3-.4. On the other hand, I might be making a distinction that is not really there and they mean the same thing.
      (2 votes)

Video transcript

- [Voiceover] So we have a graph here of the function f of x, and I'm telling you right now that f of x is going to be an exponential function. It looks like one, but it's even nicer when someone tells you that. And our goal in this video is to figure out at what x value, so when does f of x? At what x value is f of x going to be equal to negative 1/25. And you might be tempted to just eyeball it over here, but when f of x is negative 1/25, that's like right below the x-axis. If I try to eyeball it, it would be very difficult. It's very difficult to tell what value that is. It might be at three, it might be at four. I am not sure. Well, I don't want to just eyeball it, just guess it. Instead, I'm gonna actually find an expression that defines f of x because they've given us some information here. And then I can just solve for x, so let's do that. Well, since we know that f of x is an exponential function, we know it's going to take the form f of x is equal to our initial value a times our common ratio r to the xth power. Well, the initial value is straightforward enough. That's going to be the value that the function takes on when x is equal to zero. And you can even see it here, if x is equal to zero, the r to the x would just be one. And so f of zero will just be equal to a. And so what is f of zero? Well, when x is equal to zero, which essentially we're saying, where does it intersect the y-axis? We see f of zero is negative 25. So a is going to be negative 25. When x is zero, the r to the x is just one. So f of zero is going to be negative 25. We see that right over there. Now to figure out the common ratio, there's a couple of ways you could think about it. The common ratio is the ratio between two successive values that are separated by one. What do I mean by that? Well, you could view it as the ratio between f of one and f of zero. That's going to be the common ratio. Or the ratio between f of two and f of one. That is going to be the common ratio. Well, lucky for us, we know f of zero is negative 25. And we know that f of one, when x is equal to one, y, or f of x, or f of one is equal to negative five. And so just like that we're able to figure out that our common ratio r is negative five over negative 25, which is the same thing as 1/5. Divide a negative by a negative, you get a positive. So you get a five over 25, which is 1/5. We can write an expression that defines f of x. f of x is going to be equal to negative 25 times 1/5 to the x power. And so let's go back to our question. When is this going to be equal to negative 1/25? So when is this equal to negative 1/25? Well, let's just set them equal to each other. There's a siren outside. I don't know if you hear it. I'll power through. So let's see at what x value does this expression equal negative 1/25? Let's see, we can multiply. Well, actually, we want to solve for x. So let's see, let's divide both sides by negative 25. And so we are going to get 1/5 to the xth power is equal to, if we divide both sides by negative 25, this negative 25 is gonna go away. And on the right-hand side, we're going to have, divide a negative by a negative is going to be positive. It's going to be 1/625. And 1/5 to the xth power. This is the same thing as one to the xth power over five to the xth power is equal to 1/625. Well, one to the xth power is just going to be equal to one. It doesn't matter that we have this to the xth power over here. I thought I was erasing that with a black color. There you go, that's a black color right over there. So we can see that five to the xth power needs to be equal to 625. So let me write that over here. Whoops, didn't change my color. Five to the xth power needs to be equal to 625. Now the best way I could think of doing this is let's just think about our powers of five. So five to the first power is five. Five squared is 25. Five to the third is 125. Five to the fourth, we'll multiply that by five, you're going to get 625. So x is going to be 4 cause five to the fourth power is 625. So we can now say that f of four is equal to negative 1/25. And once again, you can verify that. You can verify that right over here. 1/5 to the fourth power is gonna be 1/625. Negative 25 over positive 625 is going to be negative 1/25. So hopefully that clears things up a little bit.