# Solving for constant in exponential

• Nov 14th 2008, 03:43 PM
Craka
Solving for constant in exponential
Need to solve for k in the following formula $M = M_0 2^{ - kt}$

given that Mo reduces by a factor of 16 in 32 days, the what factor will it reduce by in 10days.

I get as far as this

$
\begin{array}{l}
\frac{1}{{16}}M_0 = M_0 2^{ - kt} \\
\frac{1}{{16}} = 2^{ - kt} \\
\end{array}
$

I think i need to use logs here to solve, but I'm not sure how, as i would be base 2 I think
• Nov 14th 2008, 03:46 PM
mr fantastic
Quote:

Originally Posted by Craka
Need to solve for k in the following formula $M = M_0 2^{ - kt}$

given that Mo reduces by a factor of 16 in 32 days, the what factor will it reduce by in 10days.

I get as far as this

$
\begin{array}{l}
\frac{1}{{16}}M_0 = M_0 2^{ - k{\color{red}(32)}} \\
\frac{1}{{16}} = 2^{ - {\color{red}32}k} \\
\end{array}
$

I think i need to use logs here to solve, but I'm not sure how, as i would be base 2 I think

Note the corrections in red.

The you have $2^{-4} = 2^{ - {\color{red}32}k}$ and it should be clear what happens next.
• Nov 14th 2008, 03:57 PM
Craka
I just worked it out thanks guys.

$
\begin{array}{l}
\log _2 (\frac{1}{{16}}) = - kt \\
\log _2 (1) - \log _2 (16) = - kt \\
0 - 4 = - kt \\
- 4 = - kt \\
\end{array}
$

Is it alright to do it this way aswell, that is to take the natural log of 1/16 and divide it by the natural log of 2 to solve for -kt. As it too resulted in -4= -kt or was that just luck in this case?
• Nov 14th 2008, 04:03 PM
mr fantastic
Quote:

Originally Posted by Craka
I just worked it out thanks guys.

$
\begin{array}{l}
\log _2 (\frac{1}{{16}}) = - kt \\
\log _2 (1) - \log _2 (16) = - kt \\
0 - 4 = - kt \\
- 4 = - kt \\
\end{array}
$

Is it alright to do it this way aswell, that is to take the natural log of 1/16 and divide it by the natural log of 2 to solve for -kt. As it too resulted in -4= -kt or was that just luck in this case?

Since you still have t floating around I'm guessing you haven't bothered to read my previous reply all that closely.
• Nov 14th 2008, 04:38 PM
Craka
After i posted my question it dawned on me how to solve it. Sorry for the inconvenience it was not that I was not bothered to read your post simply that i did not see it due to the short period between my posts.
• Nov 14th 2008, 06:34 PM
mr fantastic
Quote:

Originally Posted by Craka
After i posted my question it dawned on me how to solve it. Sorry for the inconvenience it was not that I was not bothered to read your post simply that i did not see it due to the short period between my posts.

So you realise that k = 1/8, right ....?
• Nov 14th 2008, 06:53 PM
Craka
Yes thankyou. Thought the question about doing the natural log on both side and dividing through is that okay to do?
• Nov 14th 2008, 07:03 PM
mr fantastic
Quote:

Originally Posted by Craka
Yes thankyou. Thought the question about doing the natural log on both side and dividing through is that okay to do?

It's fine (except that the t should be a 32).