- 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) = A

_{0}(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 A

_{0 = }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.

4^{2x-1} = 3^{x+2}

log4^{2x-1} = log3^{x+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

- Apply the Quadratic Formula

Lastly, we learned how to solve questions like this:

2^{x} – 2^{-x} = 4

^{x}in order to generate a suitable equation:

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)

^{2x}

(2^{x})^{2} – 1 = 2^{x}(4)

And then write in standard form:

(2^{x})^{2} – 4(2^{x}) – 1 = 0

Then you apply the quadratic formula, leaving you with 2 + sqrt(5) and 2 - sqrt(5).

Alternatively, you could have set k = 2^{x }and then solved for k which would have yielded

2^{x} = 2 + sqrt(5) and

2^{x} = 2 - sqrt(5)

^{x}= 2 - sqrt(5), the answer would be less than 0, which is not possible since powers always have to be a positive number so this answer can be omitted.

That's about it. Thanks for reading.

i think this is pretty cool

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