Question 627217: After t years, the annual sales in hundreds of thousands of units of product q is given by q=(1/2)^(.8)^t
1)After how many years will the annual sales be about 95,350 units? Show your work. (Hint: You will have to take the log of both sides twice.)
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! Is it
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 ?
They are not the same. I have to assume that it is the second one because it it was the first one, you could just raise 1/2 to the 0.8 power, which is 0.57434918, and rewrite the equation as  . If you posted exactly what was given to you then the second one is correct because exponents, like the t, apply only to what is immediately in front of them! And the 0.8, not the (1/2), is what is immediately in front of it.
Also, for the sake of my convenience I am going to use 0.5 instead of 1/2.
Maybe the trickiest part of this problem is figuring out what value to use for q. You're told that q is measured in hundreds of thousands. So for a sales number of 95,350 we will need to divide it by 100,000 to find how many hundreds of thousands that is. Dividing 95,350 by 100,000 we get:
q = 0.9535
So the equation we will use to solve this problem is:
To solve an equation, with the variable in the exponent, logarithms are usually used. A logarithm of any base may be used. But there are advantages to choosing certain bases:- Choosing base that matches the base of the exponent will result in a simpler expression for an answer. (I'll show you this later.)
- Choosing a base that your calculator "knows", base 10 (log) or base e (ln), will result in a less simple expression but one that will easily be converted to a decimal approximation. (I'll start with this since I presume a decimal approximation of the answer is desired.
Finding the ln of each side:
Next we use a property of logarithms,  , to "move" the exponent of the argument out in front. (It is this very property that is the reason we use logarithms. It allows us to move the exponent, where the variable is, to a location where we can "get at it" with regular algebra.) Using this property we get:
We have made progress but the variable is still in an exponent. So we will, as the hint told you, have to use logarithms again. But before we do that, I'm going to make the side with the exponent simpler by dividing both sides by ln(0.5):
ln again:
Property again:
Divide both sides by ln(0.8):
This is an exact expression for the solution to your problem. For a decimal approximation, get out your calculator. (Note: If we had used base 10 logs, "log", all those ln's would be replaced by log's. Although the individual logarithms work out differently, the final results works out the same with either ln or log.)
If we weren't that interested in a decimal approximation, preferring a simpler expression as a result, then we would choose bases of logarithms that match the bases of the exponents:
Using a base 0.5 logarithm:
Log property:
For all bases of logarithms,  . so  . (This is how matching the bases gets us simpler expressions.) So now we have:
Using base 0.8 logarithms:
Log property:
Since  :
This is another exact expression for the solution to your problem. This is the simplest possible exact expression for the solution to your equation.
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