# Thread: How to find the Changed Base

1. ## 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)

2. 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)

3. Hello, AlphaRock!

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

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

Suppose we wish to change to base $\displaystyle b.$

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

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

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

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

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

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

4. Originally Posted by Soroban
Hello, AlphaRock!

Suppose we wish to change to base $\displaystyle b.$

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

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

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

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

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

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

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

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

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

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