Quiz 0408

 

The primary laws of logarithms include

 

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

 

1.  Solve for y the equation log(y) = 3 log(x) + .699.  First rewrite the equation as 10^left-hand side = 10^right-hand side.  Then use the laws of logarithms to simplify. 

 

2.  Solve the following equations, where it is possible to do so:

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)

 

When linearizing a certain data set log(y) vs. log(x) gives us a straight line with slope .037 and vertical intercept 2.4.  Write the corresponding equation and solve it for y.

 

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

 

When linearizing a certain data set log(y) vs. x gives us a straight line with slope -.025 and vertical intercept 12.  Write the corresponding equation and solve it for y.

 

Write the corresponding equation and solve it for y

 

 

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).

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:

 

Practice Test 1

 

 


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.   That is, log(a) + log(b) = log(a * b), so log(a+b) is not log(a) + log(b).

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.

 

We could also have solve the equation as follows:

It is worthwhile to compare this solution with the previous solution in order to see how the laws of exponents and logarithms are used slightly different ways to arrive at the same result.

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?

log(y) = .037 log(x) + 2.4. 

Note that it is important to label the axes with the quantities they actually represent; it would be a very bad idea to label the vertical axis y and the horizontal axis x, since the quantities plotted on those axes are in fact log y and log x.

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.

 

When linearizing a certain data set log(y) vs. x gives us a straight line with slope -.025 and vertical intercept 12.  Write the corresponding equation and solve it for y.

 

The equation is

 

This is an exponential function with growth factor .94, implying a growth rate of -.06.

 

Compare this solution with our previous solution of a log y vs. log x linearization.  You will see why the former gave us a power function and why this solution gives us an exponential function.

 

Generalizing, any time log y vs. x is linear we get an exponential function, and any time log y vs. log x is linear we get a power function.

 

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%?

The population function is

The population will double when P(t) reaches double its initial 1400 so the equation is

It's not required with the present statement of the problem but might be on another problem to simplify this equation, which we easily do, dividing both sides by 1400 to get

This equation can then be solved using logs, but that isn't required until the next test.

On this test you could get away with the model obtained if the population function is compounded every year rather that continuously.  This would not be correct and it's better to use the correct 'continuously compounded' model P(t) = 1400 e^(.08 t) but at this point we haven't clearly made that distinction so the function

The equation for doubling time would then be

which would simplify to

Problem Number 2

Solve using ratios instead of functional proportionalities: 

Mass is proportional to volume, which for geometrically similar objects is proportional to the cube of any given linear dimension.  In this case we conclude that mass is proportional to the height.

Thus we have a y = k x^3 proportionality, which as we have seen implies the

x2 and x1 are the diameters, y2 and y1 the masses.  We are given x2 = 7.9 m and x1 = 3.6 m, with y1 = 23328 kg and y2 unknown.  This gives us

This is the only correct way to use ratios to solve this problem.

Area is covered by little squares so the proportionality is y = k x^2 and the ratio equation is y2 / y1 = (x2 / x1)^2 so that

Problem Number 3

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

a(330) is obtained from a(1) by making 329 'jumps' each of 5 units, resulting in a total 'jump' of 329 * 5 = 1645 units.  Starting from a(1) = 7 this puts us at 7 + 1645 = 1652.  So a(330) = 1652.

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 length(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?

A graph of length vs. load has a slope that is equal to the rate of change of length with respect to load, so the slope is .05 cm / pound.

The y = m x + b form of the equation is therefore

When load is 80 cm the length is 1.58 pounds so we know that

Our final model is therefore

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.

Plug the three data points into the form y = a t^2 + b t + c and get three simultaneous linear equations in a, b and c.  Solve to get a, b and c and plug back into the form to get your model.

 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.

Set .019t^2 - 1.4 t + 40 = 0 and use the quadratic formula to determine whether there are any real solutions.  If the discriminant is positive then it has a positive square root and there will be two distinct real solutions.  If the discriminant is zero its square root is zero and there is one repeated solution.  If the discriminant is negative you cain't take its square root and you get no real solutions.

Problem Number 6

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

The vertex occurs at t = -b / (2a), as indicated by the quadratic formula.  The y coordinate of the vertex is obtained by substituting its t coordinate.  You need to give both coordinates.

The parabola opens upward since a = .028 is positive.  If the vertex is above the t axis there will be no zeros, since the parabola will never intersect the t axis.  If the vertex is on the t axis then the parabola intercepts the t axis only at this point and there is exactly one (repeated) zero.  If the vertex is below the t axis the parabola will intersect the t axis at two different points and there will be two distinct zeros.

The quadratic formula will confirm this analysis and give us the details, including whether the discriminant is positive, negative or zero and if the discriminant is not negative we will get the t values of the zeros of the function.

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?

The y coordinates of the t = x and t = x + h points are y = 2 x^2 + 11 x - 95 and y = 2 (x+h)^2 + 11 (x+h) - 95.

The rise between the points is therefore the difference 2 (x+h)^2 + 11 (x+h) - 95 - (2 x^2 + 11 x - 95) of the y coordinates.

The run is the difference x + h - x = h of the t coordinates.

The slope is therefore

We can simplify this expression to get

= (4 x h + 2 h^2 + 11 h) / h

= 4 x + 11 + 2 h.

If we let h approach zero we get a limiting value of 4x + 11.  This expresses the rate of change of the original function at t = x.

Problem Number 8

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

The form is y = m x + b.  Substituting we get the two equations

Subtracting the first equation from the second we get

Substituting back into the first we get

So our equation is

 

The graph should show the points (7, 3) and (3, 9), as well as an arbitrary point (x, y) on the graph.

The slope between the first two points is (9 - 3) / (3 - 7) = -3/2.  The graph should show the rise and the run between these points.

The rise and run between the first point and (x, y) are (y - 3) and (x - 7).  So the slope can also be expressed as (y-3) / (x-7).  Setting the two expressions for slope equal we have

This agrees with the first method.

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

The vertical coordinate is the velocity and the horizontal the clock time.  The rise therefore represents the change in the velocity in mph and the run represents the change in clock time in seconds.  So the slope = rise / run gives us change in vel / change in clock time, which we recognize as the average rate of change of velocity with respect to clock time in mph / second.

Our equation is y = -3/2 x + 27/2, which we rewrite as

 

Substituting clockTime t = 10 sec we get

 

Again writing

we substitute 14 mph for velocity to get

Subtracting 27/2 mph from both sides we have