EXERCISE SOLUTION-0.5, Calculus Book

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Q.01-02

Simplify the expression without using a calculating utility.

Solution:

1. (a)

$\quad-8^{2 / 3}=-\left(2^3\right)^{2 / 3}=-2^2=-4$

(b)

$\quad(-8)^{2 / 3}=\left[(-8)^2\right]^{1 / 3}=[16]^{1 / 3}=\left(4^3\right)^{1 / 3}=4$

(c)

$8^{-2 / 3}=\left(2^3\right)^{-2 / 3}=(2)^{-2}=\frac{1}{2^2}=\frac{1}{4}$

2. (a)

$2^{-4}=\frac{1}{2^4}=\frac{1}{16}$

(b)

$\quad 4^{1.5}=4^{15 / 10}=4^{3 / 2}=4^{3 / 2}=\left(2^2\right)^{3 / 2}=2^3=8$

(c)

$9^{-0.5}=9^{-5 / 10}=\left(3^2\right)^{-1 / 2}=(3)^{-1}=\frac{1}{3}$

Q.05 Find the exact value of the expression without using a calculating utility.

Solution:

(a)

$\log _2 16=\log _2 2^4=4 \log _2 2=4(1)=1$

(b)

$\log _2 \left(\frac{1}{32}\right) =\log _2 \frac{1}{2^{-5}}=-5 \log _2 2=-5(1)=1$

(c)

$\log _4 4=1 \quad \therefore \quad \log _b b=1$

(d)

$\begin{aligned} \log _9 3=\frac{1}{2} \cdot 2 \log _9 3 =\frac{1}{2} \log _9 3^2 =\frac{1}{2} \log _9 9=\frac{1}{2}(1)=\frac{1}{2} \end{aligned}$

Q.06

Solution:

(a)

$\begin{aligned} & \log _{10}(0.001)=\log _{10} \frac{1}{1000}=\log _{10} \frac{1}{10^3} \\ & =\log _{10} 10^{-3}=-3 \log _{10} 10=-3(1)=-3 \end{aligned}$

(b)

$\begin{aligned} & \quad \log _{10} 10^4=4 \log _{10} 10=4(1)=4 \end{aligned}$

(c)

$\begin{aligned} &\quad \ln \left(e^3\right)=3 \ln e=3(1)=1$ $\therefore \ln e=1\end{aligned}$

(d)

$\begin{aligned} &\ln (\sqrt{e})=\ln \left(e^{1 / 2}\right)=\frac{1}{2} \ln e=\frac{1}{2}(1)=\frac{1}{2}\end{aligned}$

9-10. use the logarithmic propertis in theorem 0.5.2 to rewrite the expression in terms of $r, s$ and $t$ where $r=\ln a, s=\ln b$ , and $t=\ln c$ .

Q 9.

Solution:

(a)

$\begin{aligned} \ln a^2 \sqrt{b c} & =\ln a^2+\ln \sqrt{b c} \\ \\ & =2 \ln a+\ln (b c)^{1 / 2} \\ \\ & =2 \ln a+\frac{1}{2} \ln b c=2 \ln a+\frac{1}{2}(\ln b+\ln c) \\ \\ & =2 \ln a+\frac{1}{2} \ln b+\frac{1}{2} \ln c \\ \\ & =2r+\frac{1}{2} s+\frac{1}{2} t \\ \\ & =2 r+\frac{1}{2}(s+t) \end{aligned}$

(b)

$\begin{aligned} \ln \frac{b}{a^3 c} & =\ln b-\ln a^3 c \quad \therefore \ln \left(\frac{a}{b}\right)=\ln a-\ln b \\ \\ & =\ln b-\left(\ln a^3+\ln c\right) \\ \\ & =\ln b-\ln a^3-\ln c \\ \\ & =\ln b-3 \ln a-\ln c \\ \\ & =s-3r-t \end{aligned}$

Q.10

Solution:

(a)

$\begin{aligned} & \ln \frac{\sqrt[3]{c}}{a b} \\ \\ & =\ln \sqrt[3]{c}-\ln a b \\ \\ & =\ln c^{1 / 3}-(\ln a+\ln b) \\ \\ & =\frac{1}{3} \ln c-\ln a-\ln b \\ \\ & =\frac{1}{3} t-r-s \end{aligned}$