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Thread: How to find the Changed Base

  1. #1
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    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)
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  2. #2
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    Quote Originally Posted by AlphaRock View Post
    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)
    Read this: The Change-of-Base Formula
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  3. #3
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    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} $
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  4. #4
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    Quote Originally Posted by Soroban View Post
    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 $\displaystyle 1.09^t= 2^{0.1243 t}$.
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