SOLUTION: Hello, can someone please help me with this problem: Use the properties of exponents to rewrite y=4e^{-0.5t} in the form y=a(1+r)^t or y=a(1-r)^t . Round the value of r to the nea

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Question 1155415: Hello, can someone please help me with this problem:
Use the properties of exponents to rewrite y=4e^{-0.5t} in the form y=a(1+r)^t or y=a(1-r)^t . Round the value of r to the nearest thousandth. Then find the percent rate of change to the nearest tenth of percent.
I really need to know what the percent rate of change is.
I have tried but I really just don't understand how to do it.
Please help, Thanks.

Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!


in the form or

first you need to find that makes following statement true

.....take log of both sides





...write as



.....apply exponent rule

.......

.....if log same, we have





your equation in the form is:

, }





check:
-> true

or




check:
-> not true

so, your answer is:


Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
Her answer is correct, but she used logarithms.  Maybe you aren't supposed to
use logarithms. Maybe you're not studying logarithms yet.  Anyway, here's
how without using logarithms. 

Rewrite y=4e^{-0.5t} in the form y=a(1+r)^t 

   

If those are to be identical, then regardless of what number we choose
to substitute for t, both sides of that equation must be equal.

The easiest number to substitute is 0.

   

   

Since anything raised to the zero power is 1 [With one exception, the undefined
00, which is of no concern to us here.]

So the above equation becomes:

   
     

So "a" is 4.  So we substitute 4 for a:

   

We divide both sides by 4

   

Now we substitute t=1

   

   

   

   

You can rewrite r as



which she did.

So to write 

y=4e^{-0.5t} in the form y=a(1+r)^t is to write it



And that simplifies to her answer, which is correct but not quite in
the form you asked for.  

Edwin




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