Saturday, December 26, 2009
This post is going to cover chapter 7.2, which we covered in class on Wednesday Decmber 16th 2009.
Most of the lesson was based on Radioactive decay and powers of different bases and applying quadratic formula.
Radioactive Decay is the process by which element transforms into a different element. This can be modelled by the following equation:
A(t) = A0(1/2)^t/h ,
where A(t) represents the mass of a substance, Ao represents the initial amount, t represents time, and h represents half life.
In order to solve this equation, 3 variables are needed, so you may isolate the 4th and missing variable.
For more help and for an example of this problem, check out pg. 370 of the textbook.
Powers with Different Bases
When solving powers with different bases you use an algebraic reasoning, where you take the logaritm of both sides and apply the lower law of logs to remove the varibles from the exponent.
Ex: 4^ x+1 = 64^2x
log (4^x+1) = log (64^2x)
(x+1) log 4 = 2x log 64
x+1 = 2x log 64
x+1 = 2x (3)
x+1 = 6x
1/5 = x
Applying Quadratic Formula
The third method we used to solve is using the quadratic formula
2^x – 2^-x = 4
To use the quadratic formula to solve this equatation, you must multiply both sides by 2^x so that a quadratic equation is obtained in terms of 2^x
2^x(2^x – 2^-x) = 2^x(4)
2^x(2^x) - 2^x(2^-x) = 2^x(4)
2^2x – 2^0 = 2^x(4)
2^2x – 1 = 2^x(4)
Then apply the power of a power law to the term 2^x:
(2^x)^2 – 1 = 2^x(4)
And then write in standard form (az^2+bz+c = 0)
(2^x)^2 – 4(2^x) – 1 = 0
Then you apply the quadratic formula to determine the roots.
This will result in 2 + sqrt(5) and 2 - sqrt(5).
Alternatively, you could have set b = 2^x and then solved for b which would have yielded
2^x = 2 + sqrt(5) &
2^x = 2 - sqrt(5)
At time point you must use the logarithm of both sides by using the power law of logs, and then dividing both sides by log2. This will yeild 2.08 which would be one root of this equation.
The second equation however yeilds a number smaller than 0 which results in no answer since a power must always be a positive number.
Hope that helped ! :)
Monday, December 21, 2009
This is our second last unit of our Advanced Functions course. YAYYYY ^^
In the lesson, it will be separated into two parts. The first part will talk about the exponential functions and the second part will talk about its inverse.
Since this is the first lesson of the new logarithmic unit, most of the information presented should be learned by students.
PART 1- EXPONENTIAL FUNCTIONS
In part 1, we will learn how to solve for the exponent with different bases. The "log" function on your calculators will be used greatly in this unit.
Side note: the default base of the "log" key on your calculators is ALWAYS 10.
PART 2- Inverse
In part 2, we will be pretty much doing the same thing as grade 11. Switch x and y in the same equation and isolate for X again.
Exponential functions can be represented in 3 DIFFERENT ways.
1: General Form- the simplest form of representation of an exponential equation
y=5x or y=1/5x
2. Table of Values- a table which a list of the "x" values with the corresponding "y" values
REMEMBER: the first ratio of differences of the x and y values are always constant.
How do we find the first ratio of differences?
The values of : Y2-Y1 = answer.
3: Exponential functions- this is a graph representation of the exponential functions. However, just by looking at a graph cannot help you determine whether an exponential function is or not.
Rate of Change:
To determine the rage of change of an exponential function is really much similar to finding the rate of change of any other functions.
1) Pick the x values that are really close to the value you are given to find the IROC.
2) For example: if the value that is given is 2. You may go to the left of the value or right to the value which results in x = 2.0001 or x = 1.9999.
3) Sub these values back into the original equation and find the “y” values
4) To choose the two closet points that lead to the given value will give you the best and most accurate slope of secant that is required for the answer.
5) After having the “x” values and the “y” values. Put it into the general equation used to find slopes which is:
AROC [x value – x value] = y2-y1 / x2-x1
6) To conclude your answer, you must write, “Therefore, the IROC of the given value is approximately at “answer of AROC of the 2 x values”.
The graph of the inverse function of a logarithmic function is a reflection on the y= x function.
To find an inverse function of a logarithmic function, switch x and y from the equation and solve for x again.
By switching (x,y) -> (y,x) , these can be your new inverse values that can be used to apply transformations.
Differences and similarities between an inverse function and an exponential function
1) A exponential function will have an increasing slope while the inverse function will have an decreasing slope
2) A horizontal asymptote (y = a value) is present in an exponential function. A vertical asymptote (x = a value) is present in the inverse function
3) Depending on the transformations, the horizontal asymptote or the vertical asymptote is affected by the vertical or horizontal shifts.
Sunday, December 20, 2009
- Radioactive Decay
In this chapter, we learned how to find the half-life of radioactive substances, which means we know how to find the half-life of pretty much anything that works in a similar fashion. The equation we are given is:
A(t) = A0(1/2)t/h
All we have to do now is substitute a corresponding value in each variable.
t = time, h = half-life, A(t) = mass of substance after a period of time, and A0 = mass of substance at 0 time. If you have any 3 of the 4 variables, then it is possible to isolate and solve for the unknown variable. An example is shown on page 370 of our textbooks.
- Powers With Different Bases
For these types of questions, we can either solve by Applying Algebraic Reasoning or Using a Graphing Calculator.
Applying Algebraic Reasoning:
For this method, we take the log of both sides and use the power law of logarithms.
42x-1 = 3x+2
log42x-1 = log3x+2
(2x - 1)log4 = (x + 2)log3
2xlog4 – log4 = xlog3 + 2log3
2xlog4 – xlog3 = 2log3 + log4
x(2log4 – log3) = 2log3 + log4
x = 2log3 + log4 / 2log4 – log3
After that, just use a calculator if you want to determine the approx. value of the expression. In this case, the answer is approximately 2.14.
- Apply the Quadratic Formula
Lastly, we learned how to solve questions like this:
2x – 2-x = 4
2x(2x – 2-x) = 2x(4)
2x(2x) - 2x(2-x) = 2x(4)
22x – 20 = 2x(4)
22x – 1 = 2x(4)
Then use the power of a power law to the term :
Then use the power of a power law to the term22x
(2x)2 – 1 = 2x(4)
And then write in standard form:
(2x)2 – 4(2x) – 1 = 0
Then you apply the quadratic formula, leaving you with 2 + sqrt(5) and 2 - sqrt(5).
Alternatively, you could have set k = 2x and then solved for k which would have yielded
2x = 2 + sqrt(5) and
2x = 2 - sqrt(5)
That's about it. Thanks for reading.