If f(x)f(x) is an exponential function where f(1.5)=5f(1.5)=5 and f(7.5)=79f(7.5)=79, then find the value of f(3)f(3), to the nearest hundre

Question

If f(x)f(x) is an exponential function where f(1.5)=5f(1.5)=5 and f(7.5)=79f(7.5)=79, then find the value of f(3)f(3), to the nearest hundredth.

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niczorrrr 5 days 2021-07-22T16:08:32+00:00 2 Answers 2 views 0

Answers ( )

    0
    2021-07-22T16:09:38+00:00

    Answer:

    7.93

    General Exponential Form: y=ab^x

    Plug in both points

    divide the equations

    cancel out a, subtract exponent of b

    0
    2021-07-22T16:09:50+00:00

    Answer: 7.93

    Step-by-step explanation:

    Given

    f(x) is an exponential function. Suppose f(x) is ae^{bx}

    f(1.5)=5\ \text{and}\ f(7.5)=79

    \Rightarrow 5=ae^{1.5b}\\\Rightarrow \ln 5=\ln a-1.5b\quad \ldots(i)

    Similarly,

    \Rightarrow 79=ae^{7.5b}\\\Rightarrow \ln(79)=\ln a+7.5b\quad \ldots(ii)

    Subtract (i) and (ii)

    \Rightarrow \ln (79)-\ln (5)=9b\\\Rightarrow \ln (\frac{79}{5})=9b\\\\\Rightarrow b=\dfrac{\ln (\frac{79}{5})}{9}\\\\\Rightarrow b=0.3066

    Insert the value of b

    \Rightarrow 5=ae^{0.46}\\\Rightarrow a=5\times 0.6312\\\Rightarrow a=3.156\approx 3.16

    So, the function becomes

    \Rightarrow f(x)=3.16e^{0.3066b}

    \Rightarrow f(3)=3.16e^{0.3066\times 3}\\\Rightarrow f(3)=7.927\approx 7.93

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