## Saturday, December 26, 2009

### 7.2 Solving Exponential and Log Equations

Hey Guys,
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
log 4

x+1 = 2x (3)
x+1 = 6x
1= 5x
1/5 = x

The third method we used to solve is using the quadratic formula

example:

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 ! :)