+0  
 
+3
525
24
avatar+261 

Hello all.

Once there was a giant army of mutated grizzly bear warriors. There were 100 thousand of them.

In each war 20% of them would die or settle down in villages they'd take over.

Whatever the reason would be 20% would be gone.

 

Assuming no recruitment was done, how many wars would it take for the amount of grizzly warriors to be less than 1 thousand?

 

Hope you all understood this question correctly, I'm having troubles with logarithms hope you people will be kind enough to help me! Thanks in advance!

Quazars  Jul 9, 2014

Best Answer 

 #24
avatar+394 
+25

What about my detailed answer? by Melody

*****

Just saw it sorry about that Melody was scrolling by the messages too quickly. Gave you a +5, appreciate the explanation! See the edit by Quazars

************************

The reason you missed it is the vast quantity of LANDMINE posts here.

These are grizzly bear warriors so substitute what bears do in the forest for landmines.

This basic war question was solved using logarithms and the laws of logarithms.

Calculate the logarithmic rate these landmine posts would disappear if a -5 were placed on each post that represented a land mine.

If a post asks a question about how to do something or understand it better that is one thing, if it just throws around bear p**p that is another.

OK Melody and CPhill what is the logarithm formula for reducing the amount of bear p**p landmines if a -5 is applied to those posts? Part 2 what is the logarithm formula if a -11 were applied to the posts.

7UP

SevenUP  Jul 9, 2014
 #1
avatar+4150 
+3

Three times just take 20% off of 100,000 each time until you get below a thousand which in this case took three times.

zegroes  Jul 9, 2014
 #2
avatar+261 
+3

Not correct, 0.8^3 * 100000 is not <1000

Quazars  Jul 9, 2014
 #3
avatar+4150 
+3

I was just trying to help dont give me minus five for it!

zegroes  Jul 9, 2014
 #4
avatar+261 
+3

Sorry I'm new here I thought that's what you do

Quazars  Jul 9, 2014
 #5
avatar+92781 
+3

okay.  We hardly ever give minus points here.  Zegroes was just trying to help.

So give Zegroes back his points and then I will help you.  

Melody  Jul 9, 2014
 #6
avatar+261 
+3

Already did.

Quazars  Jul 9, 2014
 #7
avatar+8258 
0

This is confusing.

DragonSlayer554  Jul 9, 2014
 #8
avatar+8258 
0

My estimation is 200.

DragonSlayer554  Jul 9, 2014
 #9
avatar+261 
+3

If it's any help I'm stuck at 100 000 x 0.8^n = 1000 where I'm trying to find n

0.8^n= 0.1 how do I calculate for n?

 

I think the answer is 21 but how would I calculate for it without using trial and error?

Quazars  Jul 9, 2014
 #10
avatar+8258 
0

What is n?

DragonSlayer554  Jul 9, 2014
 #11
avatar+261 
+3

n = amount of wars

Quazars  Jul 9, 2014
 #12
avatar+8258 
0

There are going to be 5 wars.

DragonSlayer554  Jul 9, 2014
 #13
avatar+4150 
+3

Can you please explain?i still think its three wars.

zegroes  Jul 9, 2014
 #14
avatar+92781 
+22

Zegroes, the problem with your logic is that you are taking 20% off the original amount each year.

NOT 20% of the the NEW amount.

sorry - I was called a way on a phone call. 

 

$$\begin{array}{rlll}
100000\times 0.8^n &<&1000&\\\\
100000\times 0.8^n &<&1000&\mbox{ Divide both sides by 100000}\\\\
0.8^n &<&0.01&\mbox{ When you are looking for a power you must take the log (a log IS a power)}\\\\
log0.8^n &<&log0.01&\mbox{ }\\\\
nlog0.8&<&-2&\mbox{Remember: log 0.8 is negative }\\\\
n&>&\frac{-2}{log0.8}&\mbox{ }\\\\
n&>&20.63&\mbox{ }\\\\


\end{array}$$

 

 

So it will take 21 wars.

Melody  Jul 9, 2014
 #15
avatar+261 
+3

Every war, 20% of the population dies/leaves the army. How many wars would it take for the population to drop below 1000? Original population is 100 000

0.8 since 1-20%=0.8

Quazars  Jul 9, 2014
 #16
avatar+8258 
0

There are 5 wars because 100,000*20%=20,000. Then you multiply 20,000 by 5 and your answer will be 100,000. So, there would be 5 wars.

DragonSlayer554  Jul 9, 2014
 #17
avatar+87301 
+19

Let's take it from what you have, Quazars......

We have

1000 = 100000(.8)n

Divide both sides by 100000

1/100 = (.8)n

Take the log of both sides

log(1/100) = log (.8)n

And by a property of logs, we can bring the "n" out front

log(1/100) =n log(.8)

Divide both sides by log (.8)

log(1/100) / log(.8)  = n = about 21 wars are required

CPhill  Jul 9, 2014
 #18
avatar+8258 
0

Sorry that I got the wrong answer Quazars.

DragonSlayer554  Jul 9, 2014
 #19
avatar+261 
+3

Thank you Phil for the detailed answer! Appreciate everyone trying to help as well!

 

EDIT: Sorry missed your post Melody as I was scrolling too fast!

Quazars  Jul 9, 2014
 #20
avatar+8258 
0

Quazars, please tell zegroes to answer my messages.

DragonSlayer554  Jul 9, 2014
 #21
avatar+92781 
+11

What about my detailed answer?  

 

Thanks for the points Quazars - We all like those. 

Melody  Jul 9, 2014
 #22
avatar+261 
+5

Just saw it sorry about that Melody was scrolling by the messages too quickly. Gave you a +5, appreciate the explanation! See the edit

Quazars  Jul 9, 2014
 #23
avatar+92781 
+3

Thank you.

Melody  Jul 9, 2014
 #24
avatar+394 
+25
Best Answer

What about my detailed answer? by Melody

*****

Just saw it sorry about that Melody was scrolling by the messages too quickly. Gave you a +5, appreciate the explanation! See the edit by Quazars

************************

The reason you missed it is the vast quantity of LANDMINE posts here.

These are grizzly bear warriors so substitute what bears do in the forest for landmines.

This basic war question was solved using logarithms and the laws of logarithms.

Calculate the logarithmic rate these landmine posts would disappear if a -5 were placed on each post that represented a land mine.

If a post asks a question about how to do something or understand it better that is one thing, if it just throws around bear p**p that is another.

OK Melody and CPhill what is the logarithm formula for reducing the amount of bear p**p landmines if a -5 is applied to those posts? Part 2 what is the logarithm formula if a -11 were applied to the posts.

7UP

SevenUP  Jul 9, 2014

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