# How to find the Changed Base

• Jun 17th 2009, 09:46 PM
AlphaRock
How to find the Changed Base
The formula A = P(1.09)^t is an example of exponential growth with base 1.09. Determine an equivalent continuous growth formula using base e.

Also, what if it asks me to use a different base? (eg. 2)
• Jun 17th 2009, 10:49 PM
mr fantastic
Quote:

Originally Posted by AlphaRock
The formula A = P(1.09)^t is an example of exponential growth with base 1.09. Determine an equivalent continuous growth formula using base e.

Also, what if it asks me to use a different base? (eg. 2)

• Jun 18th 2009, 02:57 AM
Soroban
Hello, AlphaRock!

Quote:

The formula $A = P(1.09)^t$ is an example of exponential growth with base 1.09.
Determine an equivalent continuous growth formula using base $e$.

Also, what if it asks me to use a different base?

Suppose we wish to change to base $b.$

Instead of $(1.09)^t$, we want $b^{kt}$, where $k$ is a constant to be determined.

We have: . $b^{kt} \:=\:(1.09)^t$

Take logs (base $b$): . $\log_b\left(b^{kt}\right) \:=\:\log_b(1.09)^t$

. . $\text{We have: }\;kt\underbrace{\log_b(b)}_{\text{This is 1}} \:=\:t\log_b(1.09) \quad\Rightarrow\quad kt \:=\:t\log_b(1.09)$

. . Hence: . $k \:=\:\log_b(1.09)$

Therefore: . $A \;=\;P\,b^{\log_b(1.09)\cdot t}$
• Jun 18th 2009, 03:45 AM
HallsofIvy
Quote:

Originally Posted by Soroban
Hello, AlphaRock!

Suppose we wish to change to base $b.$

Instead of $(1.09)^t$, we want $b^{kt}$, where $k$ is a constant to be determined.

We have: . $b^{kt} \:=\:(1.09)^t$

Take logs (base $b$): . $\log_b\left(b^{kt}\right) \:=\:\log_b(1.09)^t$

. . $\text{We have: }\;kt\underbrace{\log_b(b)}_{\text{This is 1}} \:=\:t\log_b(1.09) \quad\Rightarrow\quad kt \:=\:t\log_b(1.09)$

. . Hence: . $k \:=\:\log_b(1.09)$

Therefore: . $A \;=\;P\,b^{\log_b(1.09)\cdot t}$

Un, Soroban, since $b^{\log_b(a)}= a$, that reduces to $A= 1.09^t$, the formula we started with.

If we want to write $1.09^t$ as an exponential with base b, we have, as Soroban said, $1.09^t= b^{kt}$. Now take the natural logarithm of both sides: $ln(1.09^t)= t ln(1.09)$ and $ln(b^{kt}= kt ln(b)$ so the equation becomes $t ln(1.09)= kt ln(b)$ and $k= \frac{ln(1.09)}{ln(b)}$.

The advantage is that your calculator has a "ln" key but not a " $log_b$" key!

In particular, if b= e, then ln(b)= 1 so $k= ln(1.09)= 0.0862$ and $1.09^t= e^{t ln(1.09)}= e^{0.0862 t}$.

If b= 2, then ln(b)= ln(2)= 0.6931 so $k= \frac{ln(1.09)}{ln(2)}= \frac{0.0862}{0.6931}= 0.1243$ and $1.09^t= 2^{0.1243 t}$.