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An exponential function is a function of the form , where n > 0, n ≠ 1, and x ∈ . n is called the base.
The graph of an exponential function passes through the points (0, 1) and (1, n). It has a horizontal asymptote of y = 0 on one side, and approaches infinity on the other side. Thus, the domain will be all real numbers, and the range will be all positive numbers.
Of course, you can transform an exponential function…the graph of is related to the graph of . The domain will be the same, but the range might be a little different.
It turns out that there is one exponential base that is more important than the others. It's called the natural number (or Euler's Number)—here are two definitions.
, or .
If you use e as the base of an exponential function, then the function is called the natural exponential.
This number arises quite naturally (!) from the study of interest rates—in particular, what happens when the number of interest periods per year increases?
In general, the value (A) of an investment of P dollars at an Annual Percentage Rate of r, compounded n times per year, is . As n increases towards infinity, this formula mutates into , the value of the investment if interest is accumulated continuously.
[1.] Describe the sequence of transformations that changes the graph of into the graph of .
Just identify a, b, c and d! Multiply the y-values by 3; shift the graph 1 unit left; shift the graph 2 units down.
[2.] $20,000 is placed into an account that pays 2% interest, compounded monthly. How much will the account be worth after 10 years?
Plug in! = $24,423.99.
Note that exponential functions are one-to-one—thus, an exponential function has an inverse!
Let n > 0 and n ≠ 1. Then for any x > 0, . The function is a logarithmic function with base n (read "log base n.").
Logarithmic functions have several properties that are simply results of their inverse relationship with exponentials. There is a nice chart of this on page 468.
Since the domain of an exponential is all reals, the range of a logarithmic must be all reals.
Since the range of an exponential is greater than zero, the domain of a logarithmic must be greater than zero.
Since exponentials pass through (0, 1), logarithmics must pass through (1, 0).
Since exponentials pass through (1, n), logarithmics must pass through (n, 1).
Since exponentials have a horizontal asymptote at y = 0, logarithmics must have a vertical asymptote at x = 0.
Simple!
And, of course, you can transform them. The graph of is related to the graph of .
The most natural (!) base for logarithms is the natural number! In fact, as it turns out, this is the only base the really makes any sense…more on that later.
Logarithms with base e are written (in High School, at least) ln(x).
Common Logarithms use a base of 10, and (in High School, at least) are written log(x).
Since Natural Logarithms are the only ones that really exist, there are those who mean the natural logarithm when they write log(x).
[3.] Evaluate, without a calculator: .
The trick is to write the argument (the fraction) as the base raised to a power. Since 25 = 32, .
[4.] Evaluate: ln(π).
Easy! Enter this into your calculator: 1.145.
[5.] Describe a sequence of transformations that changes the graph of y = ln(x) into y = ln(2x - 3).
First, rewrite: y = ln(2(x - 1.5)). Divide the x-values by 2; shift the graph 1.5 units to the right.
The only logarithm key needed on a calculator is for natural log, but most calculators include one for common logs. However, no calculator (to my knowledge) has a button for logarithms of any other base—thus, we need to know how to change the base of a logarithm to one that we can use.
Equation 1 - Change of Base Equation
[6.] Evaluate: .
Use the change of base equation! = 0.873.
[7.] Expand the expression: .
.
[8.] Condense the expression: ln(x) + 3ln(y) - 5ln(z).
.
If x is in the exponent, then you need to take the logarithm of the equation.
If x is the base, you need to take a root (square, cube, etc.).
If x is inside a logarithm, then you need to use both sides of the equation as exponents of some (carefully chosen) base.
Don't forget that you can use the calculator!
[9.] Solve: 2x +1 - 1 = 31.
2x +1 - 1 = 31 ⇒ 2x +1 = 32 ⇒ ln(2x +1) = ln(32) ⇒ (x + 1)ln(2) = ln(32) ⇒ = 4.
[10.] Solve: log(x² + 1) = 1.
log(x² + 1) = 1 ⇒ x² + 1 = 101 ⇒ x² = 9 ⇒ x = ±3.
Page last validated 2010-08-15