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Math Help - 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 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}
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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 b.

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


    We have: . 1.09)^t" alt="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}.
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