Properties of exponents are related to properties of logarithms

y = log(x) is the inverse of y = 10^x

 

ln(x) should be equal to log(x) / log(e).

solve the equation log(y) = 3 log(x) + .699.

process for linearizing and obtaining an equation for the original function

a formula can look right

To see why log(ab) = log(a) + log(b)

Find these logs:

solve the equation 5^(7x-3) = 14

The equation 2^(3x - 4) + 2^(5x - 1) = 7

The equation 2^(1/x) = 7 * 2^(x+2)

log(y) vs. log(x) gives us a straight line with slope .037 and vertical intercept 2.4.

Practice Test 1

 

 

Quiz 0406

 

Tell which of the following are true and identify which are not, and explain why:

 

 

None of the following is true.  Each looks like something else that's true, but ain't.  Check with a=2, b=3, c=4 and also figure out what true statement each is mimicking.

 

 

 

 

 

Make a table of y = 10^x for x = -3, -2, -1, 0, 1, 2, 3 and then reverse columns to form a table of y = log(x).

 

Using your table find or if necessary estimate the following.  No calculators!

 

log(10^3 * 10^-2) = log(10^1) = 1.

log(10^3) = 3 and log(10^-2) = -2.  So we see that

log(10^3 * 10^-2) = log(10^3) + log(10^-2) = 3 - 2 = 1.

Lesson:  In the first line we used the laws of exponents to add the exponents of 10, giving us the final exponent of 10.  The log is the exponent of 10, so we essentially found the log by adding the exponents.

In the third line the logs of the two numbers are the exponents of 10, and we directly added the exponents to get the log of the product.

The law illustrated here is

log(a*b)  = log(a) + log(b),

which is true because the exponent of the product of two powers of the same base is the sum of the exponents, i.e., because 10^(a+b) = 10^b * 10^c.

The table quickly shows us that x = .01 or x = 10^-2.

If the log of a number is 2 then the number is 100.  If the number is x^2 then x^2 = 100 and x = 10 or x = -10.

Lesson:  In general log(x^2) = 2 * log(x).  Even more generally

log(a^b) = b * log(a), which is related to another law of exponents.

10^3 / 10^2 = 10^3-2) = 10^1, so log(10^3 / 10^2) = log(10^1) = 1.

General rule:  log(a / b) = log(a) - log(b). 

This is directly related to the way the exponents behave, as shown above for log(a*b).

 

log(10^3) * log(10^2) = 3 * 2 = 6.

This isn't directly related to the fact that 10^3 * 10^2 = 10^5.

log(a) * log(b) isn't directly related to log(a * b).

log(10^3) / log(10^2) = 3 / 2 = 1.5, not directly related to 10^3 / 10^2, which is 10.

log(a) / log(b) isn't directly related to log(a/b), or to a/b.

log(10^3) + log(10^-2) = 3 + (-2) = 1.

Note that 10^3 * 10^-2 = 10^1.

Again, log(a) + log(b) = log(a * b).

 

 

 


During the preceding class we saw that the power function y = 5 x^3 can be linearized with a log(y) vs. log(x) graph, which has slope 3 and y-intercept .699.

We will develop the ideas of logarithms and by the end of class we will see how to solve the equation log(y) = 3 log(x) + .699 for y.

 

We observe that y = log(x) is the inverse of y = 10^x, as can be verified using a calculator.

You should check these claims out using your calculator. 

The figure below shows these relationships schematically.  We note the following:

 

The base b logarithm function is the inverse of the b^x function, as indicated in the table below:

Properties of exponents are related to properties of logarithms, as indicated in the table below.

We can illustrate the law that says that log{base b}(x) = log(x) / log(b) by looking at the relationship between the base-e logarithm ln(x) and the base-10 logarithm log(x):

Since ln(x) = log{base e}(x), which by our law is equal to log(x) / log(e), our law tells us that ln(x) should be equal to log(x) / log(e).

This tells us that the function y = ln(x) is a vertical stretch by factor 2.3 of y = log(x).

We now use the laws of logarithms to solve the equation log(y) = 3 log(x) + .699.

We begin with the equation log(y) = 3 log(x) + .699, and we raise 10 to a power equal to each side of the equation to obtain the equation

10^(log y) = 10 ^ (3 log x + .699). 

Since 10^(log y) = y by the inverse-function relationship we obtain

y = 10 ^ (3 log x + .699).

Starting with y = 10 ^ (3 log x + .699) we complete the solution:

Since b^(x+y) = b^x b^y (where x and y are not related to the x and y in the equation) we have 10 ^ ( 3 log x + .699) = 10^(3 log x) * 10^.699 and the equation becomes

y = 10^(3 log x) * 10^.699.

10^.699 is just a number we can evaluate on our calculator.  10^(3 log x) = (10^(log(x) ) ^3, and 10^log(x) = x so (10^(log(x) ) ^3 = x^3 and we now have

y = x ^ 3 * 10^.699.

Evaluating 10^.699 we obtain 5 so our final equation is

y = 5 x^3.

Note that this is the equation we linearized in the previous class to obtain the equation we just solved.  So we've come full circle.  The process for linearizing and obtaining an equation for the original function was:

In this example we knew the function y = 5 x^3 to start with.  Had we not known that the data follows from this equation, we would have found the equation anyway.


We note that a formula can 'look right' but be very wrong.  Be sure to always check out your operations on logarithms with the laws.

To see why log(ab) = log(a) + log(b), first note that 10(x+y) = 10^x * 10^y, from which it follows that log(10^(x+y) ) = log(10^x * 10^y) = log(10^x) + log(10^y) = x + y.

Whatever a and b are, each can be expressed as a power of 10.  If a = 10^x and b = 10^y, which is the case for some x and some y, then log(ab) = x + y.  But since a = 10^x, log a = x and since b = 10^y, log b = y, so x + y = log a + log b.  It follows that log(ab) = log(a) + log(b).

Note that log{base b}(x) is the power of b required to get x.

 

Find these logs:

log{base 2}(16) is the power of 2 we need to get 16.  

log{base 3}(81) is the power of 3 we need to get 81.  

log{base 5}(125) is the power of 5 we need to get 125.  

log{base 4}(64) is the power of 4 we need to get 64.  

log{base 4}(32).  4^2 = 16 and 4^3 = 64.  So log{base 4}(32) is between 2 and 3.  

log{base 8}(32).  We need the power of 8 that is equal to 32.  

Suggested problems:  

log{base 25}(125)

log{base 9}(243)

log{base 16}(32)

log{base 27}(243)

log{base 2}(16) is 4.  Verify using your calculator and the identity log{base b}(x) = log x / log b:  

log{base 3}(81) is 4.  Verify using your calculator and the identity log{base b}(x) = log x / log b:  

log{base 5}(125) is 3.  Verify using your calculator and the identity log{base b}(x) = log x / log b:  

log{base 4}(64) is 3.  Verify using your calculator and the identity log{base b}(x) = log x / log b:  

log{base 4}(32) is 2.5.

log{base 8}(32).  

 

We solve the equation 5^(7x-3) = 14.

The equation 2^(3x - 4) + 2^(5x - 1) = 7 looks like it should be amenable to the same treatment as in the preceding problem.  Taking logs of both sides looks like it should bring the x terms out of the numerator.  However there is a serious pitfall in the way of this strategy.

The problem is that log(a+b) is not equal to log a + log b, as we saw earlier.  So if after writing 

log [2^(3x - 4) + 2^(5x - 1)] = log(7)

we try to say that log(2^(3x-4)) + log(2^(5x - 1) ) = log(7), we are falling into the fallacy of linearity described earlier.

It turns out that this problem cannot be solved to find an exact value of x.  We can always approximate the solution, but there is no way to find an exact solution.

The equation 2^(1/x) = 7 * 2^(x+2) can be solved by taking the log of both sides.  

The problem is continues on the next figure.

 

Be sure to understand that log 2 and log 7 are just numbers, which we can easily evaluate using calculators.

We can now see that our equation is quadratic in x, containing x^2, x and constant terms.  We will put this equation into the form a x^2 + b x + b and then we will apply the quadratic formula so find values of x which satisfy the equation.

 

 

Suppose that in linearizing a data set we find that log(y) vs. log(x) gives us a straight line with slope .037 and vertical intercept 2.4.  What is the function that models our data?

To solve the equation log(y) = .037 log(x) + 2.4 we first write the equation 10^left side = 10^right side, as shown in the second line. 

We finally see that the data is modeled by the equation y = 251 x^.037, which is the power function x^p, with p = .037, vertically stretched by factor 251.

 

 

Practice Test 1

 

Problem Number 1

What equation would you have to solve to find the doubling time, starting at t = 3, of a population that starts at 1400 organisms and grows at annual rate 8%?

Problem Number 2

Solve using ratios instead of functional proportionalities: 

If there are 3.888 billion grains of sand exposed on the surface of the first sand pile, how many grains of sand we expect to be exposed on the surface of the second?

Problem Number 3

If a(n) = a(n-1) + 5, with a(1) = 7, then what is the value of a( 330)?

Problem Number 4

If a door spring has a length that changes linearly with the weight supported as long as the spring's length is at least 44 cm, what is the linear function weight(load) that models the length, provided that the length is 80 cm when the load is 1.58 pounds, and that the length changes at a rate of .05 cm / pound?

Problem Number 5

Problem:  Obtain a quadratic depth vs. clock time model if depths of 28.057 cm, 19.79825 cm and 15.22374 cm are observed clock times t = 9.846519, 19.69304 and 29.53955 seconds.

 Problem: The quadratic depth vs. clock time model corresponding to depths of 28.057 cm, 19.79825 cm and 15.22374 cm at clock times t = 9.846519, 19.69304 and 29.53955 seconds is depth(t) = .019 t2 + -1.4 t + 40. Use the model to determine whether the depth will ever reach zero.

Problem Number 6

Find the vertex and the zeros of y = .028 t2 + -1.6 t + 79.

Problem Number 7

If y = 2 t^2 + 11 t + -95, what symbolic expression stands for the slope between the graph points for which t = x and t = x+h?

Problem Number 8

Find the equation of a line through ( 7, 3) and ( 3, 9) by each of the following methods:

If your graph represents the velocity of a coasting automobile, in mph,  vs. clock time in seconds: