Solving Exponential and Log Equations:

Hi guys! During the last couple of class periods, we have learned how to solve equations using logs. We learned two specific methods.

The First Method:

Example:        2^x = 35

Step one - we just rewrote the equation as a log:

              

log2(<the base)35 = x

Step two - divide the log of 35 (what your original equation is equal to) by the log of 2 (the base):

log 35  

 log 2   = X

the answer for this example when done out is 5.12928, but Lisa has asked us to leave the exact answer which is:    log 35  

                   log 2 

*to help you remember that the base becomes the denominator, just look at the equation rewritten as a log, and the base is lower than the number that stands for what your original equation was equal to.

The Second Method:

Example:        2^x = 35  

Step one - take the log of both sides of the equation:

log 2^x  = log 35 

Step two - use the power rule, which allows you to bring any power in the log (x) out in front of it:

xlog 2 = log 35

Then solve by...

x =  log 35  

        log 2

*remember logs are not variables

We also learned how to simplify log equations using the Properties of Logarithms (which can be seen on Jessie's blog [the one before mine..]) 

Here's an example from our homework that we went over in class:

loga (m^8n^12) ^1/4

           a^3b^5

first we use the power rule:

1/4 loga (m^8n^12)

                 a^3b^5

then we use the quotient rule:

1/4 [loga (m^8n^12) - loga (a^3b^5)]

then we used the product rule:

1/4 [loga m^8 + loga n^12 - (loga a^3 + loga b^5)]

then we used a property in the category of other, as well as the power rule:

1/4 [8 loga m + 12 loga n - (3 + 5 loga b)]

we then just multiplied the 1/4 through the rest of the equation:

2 loga m + 3 loga n - (3/4 + 5/4 loga b)]

Hope this helps!

-Timmy

               

Logarithms - 3/21/13 - Jesse

This week we covered logarithms and logarithmic functions. A logarithmic function is the inverse of an exponential function.

Here’s an example of the exponential and logarithmic function relationship:  

y = 3 x   ---  Exponential Function

x = log3Y  ---  Logarithmic Function

In each function, the base is 3. These two functions are inverse of each other, meaning you can reflect the graph of y = 3 x  across the line y = x to obtain x =  log3Y .

If you wanted to graph the inverse relation of y = 3 x , first create a t-chart and plug in x value to get out y values. This will let you graph the exponential function. To graph the logarithmic (inverse) function, simply flip the t-chart’s x and y values. Here is what the two functions look like when graphed:

Sometimes you’ll see a logarithmic function that looks like this:

y = log X

When there is no number where the base is, then it is called the Common Log, and the base is assumed to be 10, which means:

y = log X    is really just  y = log10X

There is also a button on our graphing calculators that displays ln which is called the Natural Log. It represents log base e. So, for example,   ln X    =    logeX

---

Properties of Logarithms

In class we went over the properties of logarithms. They can be found on page 331 in the book:

*Note that the Change-of-Base Formula can also be used for ln

---

Homework Problems

The homework for the weekend was pg. 322   #51-55 odd, 65, 68   and   pg. 331   #1-23 odd.

Most of the homework required the use of logarithm properties to evaluate a function. I’ll do a few problems from the pages above.

From page 322:

55). Given   log20050     find the logarithm using the change-of-base formula

log20050 becomes (log50) / (log200)  --- do the division on your calculator and the answer is 0.74

From page 331:

3). Given   log5(5 * 125)    Use the Product Rule

log55 + log5125

=

1 + 3

= 4

5). Given  logt(8y)            Use the Product Rule

Logt(8) + Logt(y)

(leave it as that because it can’t be simplified any further)

13). Given  logt(m/8)       Use the Quotient Rule

logtm  -  logt8

19). Given  logb((p2q5) / (m4b9)     Use the Power Rule and the Quotient Rule

(2logbp + 5logbq)  -  (4logbm  -  9)

March 13 (Exercises 5.3 Worksheet)

March 13

So Tuesday's class we mainly went over our homework problems on Exponential Functions. (look back at phoebes blog of 3/7-3/8 if you need to review that.)

We did problems 7, 9, 11, 13, 14, 43, 44, and 58 for homework.
In problems 7, 9 and 11 we needed to name the transformations that were needed to transform the graph of h(x)=2^x into the given.
7. f(x) = 2^x - 5 --> Vertical Shift
9. k(x) = 3(2^x) --> Vertical Stretch
11. f(x) = 2(x+2)-5 --> Horizontal Shift (to left) and Vertical Shift (5 units down)

In problems 13, and 14 we had to match each function to the graphs that were given.
13.
f(x) = a^x       --> C
g(x) = a^x+ 3   --> A
h(x) = a^(x+5) --> B

14.
f(x) = c^x     -->B
g(x) = -3c^x  ---> C
h(x) = c(x+5)   --> A
k(x) = -3c^x-2 ---> D

43.

44.

 58.

Find an exponential function that goes through these points
 Ex#1
(0, 5), (2, 45)

y = a*b^x
y = 5b^x

Plug In!! :
 45=5b^2 --> b=3

Answer: y = 5 * 3^x

Ex#2
(1, 4), (5, 1/4) --> Decay!
y = a*b^x

For (1, 4): 4= a *b^1 --> 4/b=a
For (5, 1/4): 1/4=a*b^5

plug in --> 1/4= 4/b*b^5/1 --> 1/4= 4b^4/1 -->16b^4=1 --> b^4=1/16 --> b=1/2

Answer: y= a*1/2^x (this makes sense because it's decreasing)

Nate Hansen 03.15.13. Blog Post.

Nate Hansen 03.15.13. Blog Post. 

Hi Everyone, I am blogging for just Friday, as Thursday was filled with Pi day festivities. On Friday, we went over the previous night's homework, and worked with a worksheet Lisa handed out, titled "Exponential Modeling". 

 

I will first go over some of the trickier problems from the homework that was discussed during Friday's class (as a reminder, the homework for that day can be found on the worksheet that is called "Exercises 5.2", and the problems were: #s 60,  51, 53, 54):

 

60. The figure is the graph of an exponential growth function f(x) = Pa^x.

            (a) In this case, what is P? [Hint: What is f(0)?]

            (b) Find the rule of the function f by finding a. [Hint: What is f(2)?]

The first step to solving this problem is finding the value of P. 

As the problem hints, plug in 0 for x, and the resultant value will be the initial y-value, or P (one should also note that the starting y-value is always the y-intercept).

After examining the graph of the function, we find that when x = 0, y = 4, so P = 4.

In order to complete part (b) of the problem, we will need to plug in all known values to find the value of a. After plugging in the known values, the function should look like this:

36 = 4(a)²

By solving this equation, we find that a = 3. By extension, the rule of this function is: f(x) = 4(3)^x.

54. An eccentric billionaire offers you a job for the month of September. She says that she will pay you 2¢ on the first day, 4¢ on the second day, 8¢ on the third day, and so on, doubling your pay on each successive day.

 

(a)   Let P(x) denote your salary in dollars on day x. Find the rule of the function P.

(b)   Would you be better off financially if instead you were paid $10,000 per day? [Hint: Consider P(30).]

 

For those visual learners, you can take a look at the chart below to see the growth pattern of this function:

Number of Days	Money in Dollars
0	.01
1	.02
2	.04
3	.08
4	.16
5	.32
6	.64
7	1.28
8	2.56
9	5.12
10	10.24
11	20.48
12	40.96
13	81.92
14	163.84
15	327.68
16	655.36
17	1310.72
18	2621.44
19	5242.88
20	10485.76
21	20971.52
22	41943.04
23	83886.08
24	167772.16
25	335544.32
26	671088.64
27	1342177.28
28	2684354.56
29	5368709.12
30	10737418.24

 

As shown by the above chart, the amount paid on the first day of work is .02 dollars and theoretically the value of f(0) is half of f(1) because the payment grows by a factor of 2 daily, so the value of f(0) is .01.

From this knowledge, and the fact that the function doubles every day, we find the rule of the function:

P(x) = 0.01(2)^x

To answer part (b) of this problem, we can plug in 30 for x to represent the last day of the month of September, and get a y-value of 10737418.24. Because this number and the cumulative payments for the previous days far exceeds the “$10,000 a day” figure, we know that it would make more financial sense to be paid .02 dollars on the first day of the job, and to double this amount every day afterwards for the month of September. 

The final problem I will do from Friday’s class is problem #4 on the worksheet entitled “Exponential Modeling”.

Tell whether the function represents exponential growth or decay and tell the rate of growth or decay for each.

4. j(x) = 5(4)^x÷9

Because of our knowledge of exponent rules and basic arithmetic, this problem is fairly simple. We first rewrite the problem like this:

j(x) = 5(4)^(1÷9)(x÷1)

Then, because of the “power-to-power” rule, we simplify this function so that:

j(x) = 5(4^(1÷9))^(x÷1)

In this way, we find that this function grows by 1.16652904…(simply the simplification of 4 to the one ninth), or about 17% after x time. 

 If anyone is looking for more resources to learn about Exponential Equations, check out this Khan Academy video:

Thats it for my blog post. Hope it helps. 

Nate