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Exercise 5 · Q1

Q.Find an exponential growth model, y=y0ekty=y_0e^{kt} that satisfies the stated conditions:
i. y0=1y_0=1 and doubling time t=5t=5 years.
ii. y(0)=5y(0)=5 and growth rate =2%=2\%
iii. y(1)=1y(1)=1 and y(10)=100y(10)=100

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Fit y=y0ekty=y_0e^{kt} in each case: (i) k=ln⁡25≈0.1386k=\tfrac{\ln2}{5}\approx0.1386;

(ii) y0=5, k=0.02y_0=5,\ k=0.02;

(iii) k=ln⁡1009≈0.5117, y0≈0.5995k=\tfrac{\ln100}{9}\approx0.5117,\ y_0\approx0.5995.

Exponential growth model: y=y0ekty=y_0e^{kt}, where y0y_0 = initial value (t=0t=0), kk = continuous growth rate, tt = time. Doubling: 2y0=y0ekt⇒kt=log⁡22y_0=y_0e^{kt}\Rightarrow kt=\log 2.

Part (i): y0=1y_0=1, doubling time t=5t=5 years

  1. Model: y=ekty=e^{kt}.
  2. Doubling: 2=e5k⇒5k=log⁡2⇒k=log⁡25=0.69315=0.1386.2=e^{5k}\Rightarrow 5k=\log 2\Rightarrow k=\dfrac{\log 2}{5}=\dfrac{0.6931}{5}=0.1386.
  3. Model: y=e0.1386ty=e^{0.1386t}.

Part (ii): y(0)=5y(0)=5, growth rate 2%2\%

  1. y(0)=5⇒y0=5y(0)=5\Rightarrow y_0=5; growth rate 2%⇒k=0.022\%\Rightarrow k=0.02.
  2. Model: y=5e0.02ty=5e^{0.02t}.

Part (iii): y(1)=1y(1)=1 and y(10)=100y(10)=100

  1. Write both conditions:

y0ek=1,y0e10k=100.y_0e^{k}=1,\qquad y_0e^{10k}=100.

  1. Divide the second by the first:

e9k=100  ⇒  9k=log⁡100  ⇒  k=log⁡1009=4.60529=0.5117.e^{9k}=100\;\Rightarrow\;9k=\log 100\;\Rightarrow\;k=\frac{\log 100}{9}=\frac{4.6052}{9}=0.5117.

  1. From y0ek=1y_0e^{k}=1: y0=e−k=100−1/9=e−0.5117=0.5995.y_0=e^{-k}=100^{-1/9}=e^{-0.5117}=0.5995.
  2. Model: y=0.5995 e0.5117ty=0.5995\,e^{0.5117t}.
  3. Check: y(10)=0.5995 e5.117=0.5995(166.8)≈100.y(10)=0.5995\,e^{5.117}=0.5995(166.8)\approx100. ✓
✓Final answer

  1. y=e(ln⁡2/5)t≈e0.1386ty=e^{(\ln2/5)t}\approx e^{0.1386t};
  2. y=5e0.02ty=5e^{0.02t};
  3. y=100−1/9 e(ln⁡100/9)t≈0.5995 e0.5117ty=100^{-1/9}\,e^{(\ln100/9)t}\approx0.5995\,e^{0.5117t}.

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