16=2 4. 128=2 7. In this case both 16 and 128 can be written as a power of 2. (2 4 ) 3x-5 =(2 7 ) 2x-4. Now sub the powers of 2 in for the 16 and the 128. . 12x - 20=14x - 28. A common mistake is to divide both sides by 16. DON’T DO THIS!. 16 3x-5 =128 2x-4. Exponential Equations.
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In this case both 16 and 128 can be written as a power of 2.
Now sub the powers of 2 in for the 16 and the 128.
12x - 20=14x - 28
A common mistake is to divide both sides by 16. DON’T DO THIS!
First we need the same base on each side.
Solving exponential equations requires the base on the left and right side to be equal. Consider the following example:
Now simplify the exponents on each side to get a single power of 2 there.
This creates a new equation with the same solution as the original.
We can solve this like any other equation and get the solution to the original exponential.
Now evaluate each side.
Thus x = 4 checks out.
268435456 = 268435456
Let’s apply this to a scientific example. The half-life of a substance is the time it takes for that substance to decay to one half of its original mass. The general half-life formula is given by:
Let’s check the previous question:
Where Ais the final amount of the substance, Ao is the original amount of the substance, t is the total time and h is the half-life (same units as the time)
h = 4.8d
A = 25g
t = ?
As with any question you should start with what is given.
Next write down the formula.
Now sub in all of the known values.
Consider a substance that currently has 200g. If it has a half-life of 4.8 days, how long will it take to reduce to 25g?
We want to solve for t, so we need to isolate the exponential on the right first by dividing by 200
What new equation will this create?
We need to now find a power of ½ that equals 0.125.
Now sub it in
Now that the bases are equal, we can write down a new equation that has the same solution as the old one.
Solving for t, we get our answer.
Finish the question off with a statement.
Thus it would take 14.4 days for 200g of this material to decay to 25g.
Cobalt-60 which has a half-life of 5.3 years, is used extensively in medical radiology. The amount left at any given time is given by:
a) What fraction of the initial amount will be left after 15.9 years?
b) How long will it take until there is only 6.25% of the original amount left?