# Thread: Solving for constant in exponential

1. ## 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

2. 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.

3. 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?

4. 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.

5. 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.

6. 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 ....?

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

8. 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).