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This is the second of our many quizzes on Exponential, logarithmic, and logistic functions. Click on this to access more such competitive quizzes on Algebra.

Exponential, logarithmic, and logistic functions Quiz 2

mathematics

 Shivani is learning to simplify equations, she tries simplifying the exponential term ${{({{x}^{\frac{2}{3}}}({{x}^{\frac{-3}{2}}}))}^{\frac{8}{3}}}$. What will be her answer?

 Use the identities $({{a}^{m}})({{a}^{n}})={{a}^{m+n}}$ and ${{({{a}^{m}})}^{n}}={{a}^{m\times n}}$ to solve the problem.

  ${{({{x}^{\frac{2}{3}}}({{x}^{-\frac{3}{2}}}))}^{\frac{8}{3}}} \\ $ Solving the inner bracket and adding the exponents. Using the identity, $ \\ ({{a}^{m}})({{a}^{n}})={{a}^{m+n}} \\ \Rightarrow {{({{x}^{-\frac{5}{6}}})}^{\frac{8}{3}}} \\ $ Using the identity $(a^m)^n = a^{m\times n} \\ \Rightarrow x^{{-\frac{3}{6}} \times {\frac{8}{3}}}={x}^{-\frac{4}{3}}$

 Sunanya had written an equation with its value equated to $2048$. The piece of paper went missing and she can remember that the equation was of the form ${{2}^{2x-3}}$ . What will be the missing value of $x$?

 The equation will be of format ${{2}^{2x-3}}=2048={{2}^{11}}$. Use the identity $({{a}^{m}})=({{a}^{n}})\to m=n$.

  Since the value can be equated to $2048$ $ \\ {{2}^{2x-3}}=2048={{2}^{11}} \\ $ Using the identity $ \\ ({{a}^{m}})=({{a}^{n}})\to m=n \\ \Rightarrow 2x-3=11 \\ \Rightarrow 2x=11+3 \\  \Rightarrow x=\dfrac{14}{2} \\  \Rightarrow x=7 $

 Subtract $\log (4-x)$ from $\log (6x)$, if the value is equal to $\log (3)$, then find the value of $x$.

 Use the equation $\log (a)-\log (b)=\log \left(\dfrac{a}{b}\right)$ and $\log (a)=\log (b)\to a=b$.

$\log (a)-\log (b)=\log \left(\dfrac{a}{b}\right) \\ \Rightarrow \log (6x)-\log (4-x)=\log \left(\dfrac{6x}{4-x}\right) \\ $ Given the value is equal to $\log (3)$ $ \\ $ Using the equation $ \\ \log (a)=\log (b)\to a=b \\  \Rightarrow \dfrac{6x}{4-x}=3 \\  \Rightarrow 6x=3(4-x) \\  \Rightarrow 6x=12-3x \\  \Rightarrow 9x=12 \\  \Rightarrow x=\dfrac{4}{3}$

 A psychological experiment is conducted between $1200$ people. Four participants are given a piece of information and are said to spread the information. It is observed that after three days, $300$ people know about the details. What can you say about the data for the fourth day if you go by the trends?

 The carrying capacity will be $c=1200$ and the data can be expressed as two points $(0,4),(3,300)$.

 Since the count of participants is fixed hence, we can use logistic function. The carrying capacity will be $c=1200$ $ \\ $ The data can be expressed as two points $(0,4),(3,300)$ $ \\ $ Using first point to solve it $ \\ \Rightarrow \dfrac{1200}{1+a.{{b}^{0}}}=4 \\  \Rightarrow \dfrac{1200}{4}=1+a \\  \Rightarrow a=299 \\ $ Substituting the value of a and keeping the second point to solve for $b$ $ \\  \Rightarrow \dfrac{1200}{1+299.{{b}^{3}}}=300 \\  \Rightarrow 4=1+299{{b}^{3}} \\  \Rightarrow b\approx 0.215 \\ $  Using the equation at $x=4$  $ \\ f(x)=\dfrac{1200}{1+299\times {{0.215}^{x}}}=f(4)\approx 729\text{people} \\ \log (a)-\log (b)=\log \left(\dfrac{a}{b}\right) \\ \log (6x)-\log (4-x)=\log \left(\dfrac{6x}{4-x}\right) \\ $ Given the value is equal to $\log (3)$ $ \\ $ Using the equation $ \\ \log (a)=\log (b)\to a=b \\ \Rightarrow \dfrac{6x}{4-x}=3 \\  \Rightarrow 6x=3(4-x) \\  \Rightarrow 6x=12-3x \\  \Rightarrow 9x=12 \\  \Rightarrow x=\dfrac{4}{3} $

 Suppose you have to make a logistic equation with the carrying capacity as $12$. It is given that two points that satisfy this equation are $(0,9)$ and $(1,11)$. Find the value of $a$ and $b$.

 Use the points $(0,9)\text{ and }(1,11)$ to find the general form of the equation.

 Using the first point to solve it $ \\ \dfrac{12}{1+a.{{b}^{0}}}=9 \\  \Rightarrow \dfrac{12}{9}=1+a \\  \Rightarrow a=\dfrac{1}{3} \\ $ Substituting the value of a and keeping the second point to solve for $b$ $ \\ \Rightarrow \dfrac{7}{1+(2.5).{{b}^{3}}}=5 \\  \Rightarrow 1+(2.5).{{b}^{3}}=\dfrac{7}{5} \\  \Rightarrow b\approx 0.54 \\ $ Using the equation at $x=4$  $ \\ f(x)=\dfrac{7}{1+2.5\times {{0.54}^{x}}}$ 