- 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.