Lesson Objectives:- Find common logarithms and natural logarithms without a calculator.
- Convert between exponential equations and logarithmic equations.
- Change logarithmic bases.
- Graph logarithmic functions.
- Solve applied problems involving logarithmic functions.
Graph `x = 2^y`.
We can find ordered pairs that are solutions by choosing values of y, and computing the corresponding values of x. Start by drawing a table with your x and your y-values.
And now let's start with -3, -2, -1, 0, 1, 2, and 3. And we're going to plug these values in for y to get our x-value. So for the first one, we have `2^-3` or `1/8`. For the next one we've got `2^-2` or `1/4`. And then we have `2^-1` or `1/2`. `2^0`, which is just 1. Then `2^1 = 2`; `2^2 = 4`; `2^3 = 8`.
Plot the points and complete the graph. So we have `(1/8, -3)`, `(1/4, -2)`, and `(1/2, -1)`. Then (1, 0), (2, 1), (4, 2), and (8, 3).
So our graph looks like this. This is the same as the graph of `y = log_(2)x`.
Logarithm function with base a.
The logarithm of a number is the exponent to which a fixed number must be raised to produce the given number.
Thus, `y = log_(a)x` and `x = a^y`, where `x > 0`, `a > 0`, and `a != 1` are two ways of expressing the same relationship between x and y.
Properties of Logarithms.
The logarithm of 1 is 0. This is because `a^0 = 1`, so `0 = log_(a)1`.
The logarithm of the base itself is 1. Since `a^1 = a`, `1 = log_(a)a`.
Convert to a logarithm equation.
`27^(1/3) = 3`. So this is equivalent to `log_(27)3 = 1/3`.
The logarithm of a number is the power of some base (in this case, 27), which will produce the number.
Convert to an exponential equation.
`log_(7)7 = 1`. So `7^1 = 7`.
Two Important Systems:
Although the possible number of different systems of logarithms is unlimited, there are only two systems in common use. This is the Common System and the Natural System.
The Common System has a base of 10, it's written `log_(10)x`, and can also just be written `logx`.
The Natural System has a base of e. It can be written `log_(e)x` or also be written `ln x`.
e represents the sum of the series, `1+1/1+1/(1*2)+1/(1*2*3)+1/(1*2*3*4)`, etc., as the number of terms is infinitely increased.
The value e, carried to seven places of decimals is `e = 2.7182818`, etc..
Find the following using a calculator.
`log 523`. So look for the log key, or if you're using a TI-89, you'll use your alpha keys to type in log, and then 523. So `log523` gives us 2.7185.
And then for `ln 809.4`, we're going to look for our "natural log" key, which is just "ln". And then type in 809.4. So this gives us 6.6963.
The Change of Base Formula.
When the logarithm of a number to the base a is multiplied by `log_(b)a`, we obtain the logarithm of that number to the base b.
So here we have `log_(b)y = log_(a)y*log_(b)a`.
This enables us to change from one base to another.
We'll start with the identity, `a = b^(log_(b)a)`. So `a^x = b^(log_(b)a^x)`. And since the logarithm of a power of a number is the power times the logarithm of the number, we can bring the power down, so we have `a^x = b^(x*log_(b)a)`.
Then if `y = a^x`, which is equal to `b^(x*log_(b)a)`, then by the definition of logarithms, we can write this as `log_(a)y = x`. And then for the second part, `log_(b)y = x*log_(b)a`.
And then if we plug all of this in for x, we have `log_(b)y = log_(a)y * log_(b)a`, which is our Change of Base Formula. And you may also see this as `log_(a)y = (log_(b)y)/(log_(b)a)`.
Find the logarithm using natural logarithms and the change of base formula.
`log_(4)16`. So according to the Change of Base Formula, this is equivalent to `(ln16)/(ln4)`, which we can type into our calculator, and we get 2.
Graph the function and give the domain and vertical asymptote.
`y = 1-ln(x)`. This is the graph of `y = ln(x)` reflected across the x-axis and shifted up 1 unit.
We can also create a table using our calculator. At 0, it's undefined. At 0.5, we have 1.69 for our y-value. At 1, we have 1. 1.5, we have 0.59. At 2, we have 0.3.
So for our first point, we'll plot (0.5, 1.69). And then (1, 1), (1.5, 0.59), (2, 0.3).
So that our graph looks like this when we complete it. The domain is `(0, oo)`, and we have a vertical asymptote at `x = 0`.
On March 11, 2011, an earthquake with intensity `10^(9.0)I_0` struck off the northeast coast of Japan, about 250 miles from Tokyo. What was the magnitude of this earthquake on the Richter Scale?
`R = log(I/I_0)`, where R is the magnitude of the earthquake on the Richter Scale, I is the intensity of the earthquake, and `I_0` is the minimum intensity that can be recorded on the seismograph; it's used for comparison.
So we're given `I = 10^(9.0)I_0`, and we just plug this n for I. So we have `R = log(10^(9.0)*I_0)/(I_0)`. This gives us `R = log 10^(9.0)`. And we can bring our power out, so we have `R = 9.0*log 10`. And `log 10` is the same as log base 10 of 10, which is just 1.
The magnitude of this earthquake on the Richter Scale was 9.0.