Question 1159949: Hello
The answer to the following question is 26 but I'm trying to figure out how to set up the equation and enter it into the calculator with x being the exponent.
Appreciate your help on this one. Thanks
Picture a cat stalking a mouse. They’re about 100 inches apart. Every time the mouse starts nibbling at the hunk of cheese, the cat takes advantage of the mouse’s distraction and creeps closer by one-tenth the distance between them. The cat want's to get about 6 inches away. How long will it take before the cat can pounce on the mouse.
Found 2 solutions by jim_thompson5910, ikleyn:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
x = number of nibbles the mouse makes
y = distance between the cat and mouse
x is a positive whole number, while y is a positive real number
x cannot be a fraction or decimal, but y can
When x = 0, the value of y is y = 100. This is the starting distance.
When x changes to x = 1, the value of y becomes y = 90. We take 1/10 of 100 to get 10, so the cat has moved 10 inches closer leaving 100-10 = 90 inches left til the cat reaches the mouse.
When we get to x = 2, the cat moves (1/10)*90 = 90/10 = 9 inches. So there's 90-9 = 81 inches left.
You can keep this going until you have y be 6 inches. If you can't land exactly on y = 6, then try to get as close as possible. Make sure to have y be smaller than 6, rather than larger, so that we meet our goal.
--------------------------------------------------------
As you can probably see, it will take a while to get y to 6 or smaller following the method outlined above. We can use algebra to help speed things up.
Notice that each time the cat moves 1/10 of the distance closer, there's 9/10 of the distance left to go. What we can do is multiply 9/10 by the old distance remaining to get the new distance remaining
new distance remaining = (9/10)*(old distance remaining)
So for instance, we start off with 100 and go to 90
new distance remaining = (9/10)*(old distance remaining)
new distance remaining = (9/10)*(100)
new distance remaining = 90
and this repeats for the next iteration as well
new distance remaining = (9/10)*(old distance remaining)
new distance remaining = (9/10)*(90)
new distance remaining = 81
and so on. When we have repeated multiplication, we turn to exponents. You probably already know this considering you mentioned exponents in your post.
The equation would be
y = 100(9/10)^x
which in decimal form is
y = 100(0.9)^x
Let's try x = 0 and see what happens:
y = 100(0.9)^x
y = 100(0.9)^0
y = 100(1)
y = 100
so we have x = 0 and y = 100 pair up as expected
Now try x = 1
y = 100(0.9)^x
y = 100(0.9)^1
y = 100(0.9)
y = 90
That works as well. The cat is y = 90 inches away after the mouse does x = 1 nibble.
Lastly lets try out x = 2
y = 100(0.9)^x
y = 100(0.9)^2
y = 100(0.81)
y = 81
This works too.
--------------------------------------------------------
We want to solve the exponential equation for x when y = 6
We will use logarithms to help isolate the exponent. One way you can remember is thinking "the exponent is in the trees, so we have to log it down"
Let's do so
y = 100(0.9)^x
6 = 100(0.9)^x
100(0.9)^x = 6
(0.9)^x = 6/100
(0.9)^x = 0.06
Log[ (0.9)^x ] = Log[0.06] .... apply logs to both sides
x*Log[0.9] = Log[0.06] .... see note below
x = Log[0.06]/Log[0.9]
x = 26.7027045112137
note: I'm using the log rule log(x^y) = y*log(x), which is the sole reason why we use logarithms. This rule allows us to pull the exponent down.
If x = 26, then,
y = 100(0.9)^x
y = 100(0.9)^26
y = 100(0.06461081889227)
y = 6.46108188922668
which rounds to y = 6 when we round to the nearest whole number.
I would argue that x = 26 nibbles is 1 too few because y = 6.46 is over 6 inches. The cat would probably want to get under y = 6 (if the cat can't exactly land on y = 6 itself) so I would argue x = 27 is a better fit.
y = 100(0.9)^x
y = 100(0.9)^27 ... try out x = 27
y = 100(0.05814973700304)
y = 5.814973700304
which is now under 6
However, I think I see why your teacher opted to go with x = 26 instead of x = 27.
Answer by ikleyn(52832)  (Show Source):
You can put this solution on YOUR website! .
In my view, this problem, as worded and formulated, is defective.
Indeed, it says "every time when . . . ", but does not specify what is the time interval between two adjacent events.
So, it is not possible to answer literally "how long it will take".
We only can answer, after how many discrete time intervals it will happen, but the duration
of a real "elementary" time interval remains unknown.
Any Physicist will see this deficiency MOMENTARILY . . .
/\/\/\/\/\/\/\/
Surely, I found the links in the Internet to the PRECISELY CORRECT formulation of this problem
https://books.google.com/books?id=zYBi6cxYCVkC&pg=PA60&lpg=PA60&dq=Every+time+the+mouse+starts+nibbling+at+the+hunk+of+cheese,+the+cat+takes+advantage&source=bl&ots=4j2i8imMve&sig=ACfU3U3m-TC0SSztsn6eRARM6U7NPFtyfw&hl=en&sa=X&ved=2ahUKEwiUltzA6c_pAhUIqZ4KHaQoD28Q6AEwAHoECAkQAQ#v=onepage&q=Every%20time%20the%20mouse%20starts%20nibbling%20at%20the%20hunk%20of%20cheese%2C%20the%20cat%20takes%20advantage&f=false
and
https://books.google.com/books?id=zYBi6cxYCVkC&pg=PA60&lpg=PA60&dq=Every+time+the+mouse+starts+nibbling+at+the+hunk+of+cheese,+the+cat+takes+advantage&source=bl&ots=4j2i8imOxg&sig=ACfU3U0STbrKsQ6C1Qs-q7eyCkHuVWmgNA&hl=en&sa=X&ved=2ahUKEwiEk9Wq6s_pAhUDOn0KHZU-AHgQ6AEwAHoECAoQAQ#v=onepage&q=Every%20time%20the%20mouse%20starts%20nibbling%20at%20the%20hunk%20of%20cheese%2C%20the%20cat%20takes%20advantage&f=false
I hate it very much when people bring "dirty" formulated problems to the forum.
|
|
|
