# Thread: Exponential Growth and Decay

1. ## Exponential Growth and Decay

I;m having trouble with some questions on exponential growth. All of them seem to end with the same type of equation, one that I cannot figure out.

They will; end something like this..

2=1.15^x (1.15 to the power of x)

On simpler questions like this, you had to find a common base.
ie. 27=3^x. 3^3=3^x therefore x=3

But I don't see how you can find a common base wth 2 and 1.15, or other such numbers (where one has many decimal places-for example, one question ended up as 9=1.076842^x).

The question that ended with 2=1.15^x was as follows: A bacteria grows at 15% per day. In how many days will it have doubled?

Anybody have any ideas?

2. Originally Posted by Piggins
I;m having trouble with some questions on exponential growth. All of them seem to end with the same type of equation, one that I cannot figure out.

They will; end something like this..

2=1.15^x (1.15 to the power of x)

On simpler questions like this, you had to find a common base.
ie. 27=3^x. 3^3=3^x therefore x=3

But I don't see how you can find a common base wth 2 and 1.15, or other such numbers (where one has many decimal places-for example, one question ended up as 9=1.076842^x).

The question that ended with 2=1.15^x was as follows: A bacteria grows at 15% per day. In how many days will it have doubled?

Anybody have any ideas?
$\displaystyle 2 = 1.15^x$

Take the log of both sides. It doesn't matter which base you choose. (Though I'd choose either base 10 or base "e", since you can do those directly on your calculator.) My preference is to use ln:
$\displaystyle ln(2) = ln \left ( 1.15^x \right )$

$\displaystyle ln(2) = x \cdot ln(1.15)$

$\displaystyle x = \frac{ln(2)}{ln(1.15)}$
which you can plug into your calculator.

-Dan

3. Originally Posted by topsquark
$\displaystyle 2 = 1.15^x$

Take the log of both sides. It doesn't matter which base you choose. (Though I'd choose either base 10 or base "e", since you can do those directly on your calculator.) My preference is to use ln:
$\displaystyle ln(2) = ln \left ( 1.15^x \right )$

$\displaystyle ln(2) = x \cdot ln(1.15)$

$\displaystyle x = \frac{ln(2)}{ln(1.15)}$
which you can plug into your calculator.

-Dan
Is there any other way to do it? We have not learned logs, I believe that is Grade 12 math.

4. Originally Posted by Piggins
Is there any other way to do it? We have not learned logs, I believe that is Grade 12 math.
No i can't think of another way. I could explain the basics of logs to you, if you want.

Say we have the following:

$\displaystyle 10^{x} = 100$

That can be expressed as:

$\displaystyle x = log_{10} 100$

And to solve it on your calculator, simply type $\displaystyle log 100$ divided by $\displaystyle log 10$

If the base of the log is not mentioned, it is assumed to be 10.

For example:

$\displaystyle log 2 = x$

Thus $\displaystyle log_{10} 2 = x$

Thus $\displaystyle \frac{ log 2 }{ log 10 } = x$

5. Originally Posted by Piggins
Is there any other way to do it? We have not learned logs, I believe that is Grade 12 math.
The only other way I can think of is to estimate it. For example:
$\displaystyle 1.15^4 \approx 1.74901$

$\displaystyle 1.15^5 \approx 2.01136$

So maybe try
$\displaystyle 1.14^{4.7} \approx 1.92877$

etc.

-Dan

6. Thanks for the help everyone.

1) Estimating was also the only method I could think of.

2) Like 10 people asked about it today, so the teacher explained the basics of logs to us, so now I would do it as explained by you guys.