# simple problem

• Jun 12th 2007, 10:26 PM
chinese_man_77
simple problem
i finished about all of this math problem except for the ending.
i reduced the problem to 5000^-10k

what i need is how to find the value of k
• Jun 12th 2007, 10:27 PM
Jhevon
Quote:

Originally Posted by chinese_man_77
i finished about all of this math problem except for the ending.
i reduced the problem to 5000^-10k

what i need is how to find the value of k

we need an equation or something, you haven't given enough information
• Jun 12th 2007, 10:31 PM
chinese_man_77
A = (ax)(10)^-kt

ax = 500
t= 10
A = 450

i got it down to 450 = 5000^-10k
• Jun 12th 2007, 10:31 PM
qbkr21
Re:
RE:

Even if you set this equation equal to zero there still isn't a solution...

-qbkr21
• Jun 12th 2007, 10:33 PM
chinese_man_77
its a regents question, so im pretty sure there is a solution and why would it be equal to 0 when its equal to 450
• Jun 12th 2007, 10:37 PM
chinese_man_77
A = (ax)(10)^-kt

ax = 500
t= 10
A = 450

i got it down to 450 = 5000^-10k
i just dont know how to finish off the problem
• Jun 12th 2007, 10:37 PM
Jhevon
Quote:

Originally Posted by chinese_man_77
A = (ax)(10)^-kt

ax = 500
t= 10
A = 450

i got it down to 450 = 5000^-10k

we have $450 = 500 \cdot 10^{-10k}$

$\Rightarrow \frac {450}{500} = 10^{-10k}$

$\Rightarrow \log \left( \frac {9}{10} \right) = -10k$ ........do you understand this step?

$\Rightarrow k = \frac {\log \left( \frac {9}{10} \right)}{-10}$
• Jun 12th 2007, 10:38 PM
chinese_man_77
nice, thanks a lot
• Jun 12th 2007, 10:40 PM
chinese_man_77
Quote:

Originally Posted by Jhevon
we have $450 = 500 \cdot 10^{-10k}$

$\Rightarrow \frac {450}{500} = 10^{-10k}$

$\Rightarrow \log \left( \frac {9}{10} \right) = -10k$ ........do you understand this step?

$\Rightarrow k = \frac {\log \left( \frac {9}{10} \right)}{-10}$

one more quick question... i see how you got to the end of the problem, but is there any reason why you knew it was a log problem?
• Jun 12th 2007, 10:50 PM
Jhevon
Quote:

Originally Posted by chinese_man_77
one more quick question... i see how you got to the end of the problem, but is there any reason why you knew it was a log problem?

the unknown was in the power. when we see that, we either need logs or the laws of exponents to solve the problem. logs was the appropriate choice here