Exercises 3

This page allows you to practice some exercises on Functions of One Variable. If you notice that you have difficulties, we advise you to go over the corresponding sections in the book.

These questions correspond to the following sections: - Section 4.7 - Polynomials - Section 4.8 - Power Functions - Section 4.9 - Exponential Functions - Section 4.10 - Logarithmic Functions - Extra: Trigonometric Functions

Question 18

Find all integer roots of the following equations.

(a) $x^4 - x^3 - 7x^2 + x + 6 = 0$

(b) $2x^3 + 11x^2 - 7x - 6 = 0$

(c) $x^3 - x^2 - 25x + 25 = 0$

(d) $x^5 - 4x^3 - 3 = 0$

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(a) All possible integer roots must be factors of the constant term, $6$. Since $\pm 1, \pm 2, \pm 3$ and $\pm 6$ are the factors of $6$, they are possible integer solutions. Direct substitution into the left-hand side of the equation reveals that $−2, −1, 1, 3$ are the integer roots we are after. Since a fourth-degree polynomial can have at most four roots, we have found all of them.

(b) $1, −6$.

(c) $1, 5,$ and $−5$.

(d) $−1$.

Question 19

Perform the following divisions. (a) $(2x^3 + 2x - 1) \div (x-1)$

(b) $(x^5 - 3x^4 + 1) \div (x^2 + x + 1)$

(c) $(x^4 + 3x^2 + 5) \div (x -c)$, and show that it leaves a reminder for all values of $c$.

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(a) $2x^2 + 2^x + 4 + 3/(x − 1)$.

(b) $x^3 − 4x^2 + 3x + 1 − 4x/(x^2 + x + 1)$.

(c) $c^4 + 3c^2 + 5 \geq 5 \neq 0$ for every choice of $c$, so the division has to leave a remainder.

Question 20

Solve the following equations for $t$.

(a) $2^{2t} = 8$

(b) $3^{3t + 1} = 1/81$

(c) $10^{t^{2} -2t + 2} = 100$

(d) $3^{5t}9^{t} = 27$

(e) $9^{t} = (27)^{1/5}/3$

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(a) $2^3 = 8$, so $2t = 3$ yields $t=3/2$.

(b) $3^{-4} = 1/81$, so $3t + 1 = -4$ yields $t = -5/3$.

(c) $10^2 = 100$, so we get $t^2 - 2t + 2 = 2$. This means $t^2 - 2t = 0$, which results in $t=0$ or $t=2$.

(d) $3^{5t}9^{t} = 3^{5t}(3^2)^t = 3^{5t+2t} = 3^{7t}$ and $27 = 3^3$ , so $7t = 3$, and then $t = 3/7$.

(e) $9^t = (3^2)^t = 3^{2t}$ and $(27)^{1/5}/3 = (3^3)^{1/5}/3 = 3^{3/5} /3 = 3^{−2/5}$ , and then $2t = −2/5$, so $t = −1/5$.

Question 21

A savings account with an initial deposit of €100 earns 12% interest per year. What is the amount of savings after $t$ years?

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The amount of savings after $t$ years is $100 (1 + 12/100)^t = 100 \cdot (1.12)^t$.

Question 22

Which of the following equations do not define exponential functions of $x$?

(a) $y = 3^x, \quad$(b) $y = x^{\sqrt{2}}, \quad$(c) $y = (\sqrt{2})^{x}, \quad$ (d) $y = x^{x}, \quad$(e) $y = (2.7)^{x}, \quad$ (f) $y = 1/2^{x}$

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(b) and (d) do not define exponential functions. Note that in (f): $y = (1/2)^x$ .

Question 23

Express the following as multiples of $\ln 3$.

(a) $\ln 9 \quad$(b) $\ln \sqrt{3} \quad$(c) $\ln \sqrt[5]{3^2}, \quad$(d) $\ln(1/81)$

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(a) $\ln 9 = \ln 3^2 = 2\ln 3$.

(b) $\ln \sqrt{3} = \ln 3^{1/2} = (1/2)\ln 3$.

(c) $\ln \sqrt[5]{3^2} = \ln 3^{2/5} = (2/5)\ln 3$.

(d) $\ln(1/81) = \ln 3^{-4} = -4 \ln 3$.

Question 24

Solve the following equations for $x$.

(a) $\ln x = 3$

(b) $\ln(x^2 - 4x +5) = 0$

(c) $\frac{x\ln(x+3)}{x^2 + 1} = 0$

(d) $3^{x}4^{x+2} = 8$

(e) $4^{x} - 4^{x-1} = 3^{x+1} - 3^{x}$

(f) $\log_{x} e^{2} = 2$

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(a) Note that $e^{\ln x} = x$, so $x = e^{3}$.

(b) Since $\ln 1 = 0$, we get $x^{2} - 4x + 5 = 1$, which yields $(x-2)^{2} = 0$. Thus, $x=2$.

(c) Note that the denominator can never be zero. Thus, this expression can only equal zero when $x=0$ or $\ln (x+3) = 0$. Since $\ln 1 = 0$, we have to solve $x + 3 = 1$, which means that $x= -2$. Hence $x =0$ and $x=-2$ are the two solutions.

(d) $3^{x}4^{x+2} = 8 \iff 3^{x} (2^2)^{x+2} = 2^{3} \iff 3^{x}2^{2x+4} = 2^{3}$. Now $3^{x} = 2^{3 - 2x -4} = 2^{-2x-1}$. Now, $x\ln 3 = -2x \ln 2 - \ln 2 \iff x (\ln 3 + 2\ln 2) = -\ln 2$. Thus, we obtain$x = -\ln 2/(\ln 3 + 2\ln 2) = -\ln2/\ln 12$, as $\ln 12= \ln(3 \cdot 2^2)= \ln3 + 2 \ln 2$.

(e) $4^{x} - 4^{x-1} = 4^{x} \left( 1 - \frac{1}{4} \right)$ and $3^{x+1} - 3^{x} = 3^{x}(3 - 1)$. Thus $\frac{3}{4} \cdot 4^{x} = 2 \cdot 3^{x}$. Now, we can write $4^{x}/3^{x} = (4/3)^{x} = 8/3$. Thus, by taking the natural logarithm, $x = \ln(8/3)/\ln(4/3)$.

(f) Recall that: $\log_{⁡b}(a)=c$ means $b^c=a$. Apply this to the given equation: $\log_{⁡x}(e^2)=2 \implies x^2 = e^2$. This solves for $x=\pm e$, however since the logarithm cannot have a negative base, we only have $x=e$ as a solution.

Question 25

Assume that all the variables in the formulas below are positive. Which of these formulas are always true, and which are sometimes true?

(a) $(\ln A)^{4} = 4 \ln A \qquad$(b) $\ln B = 2 \ln \sqrt{B} \qquad$(c) $\ln A^{10} - \ln A^{4} = 3\ln A^{2} \qquad$

(d) $\ln\frac{A+B}{C} = \ln A + \ln B - \ln C \qquad$(e) $\ln\frac{A+B}{C} = \ln(A+B) - \ln(C) \qquad$(f) $\ln\frac{A}{B} + \ln\frac{B}{A} = 0$

(g) $p \ln(\ln A) = \ln(\ln A^{p}) \qquad$(h) $p \ln(\ln A) = \ln(\ln A)^{p} \qquad$(i) $\frac{\ln A}{\ln B + \ln C} = \ln A (BC)^{-1}$

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(a) False. (Let $A = e$.)

(b) True. $2 \ln \sqrt{B} = 2 \ln B^{1/2} = 2(1/2) \ln B = \ln B$.

(c) True. $\ln A^10 − \ln A^4 = 10 \ln A − 4 \ln A = 6 \ln A = 3 \cdot 2 \ln A = 3 \ln A^2$.

(d) Wrong. (Put $A = B = C = 1$.)

(e) Correct. General rule of logarithms.

(f) Correct. $\ln(A/B) + \ln(B/A) = \ln A - \ln B + (\ln B - \ln A) = 0$.

(g) Wrong. (If $A = e$ and $p = 2$, then the equality becomes $0 = \ln 2$.)

(h) Correct. Note that $\ln A$ can just be seen as a number, say $x$. Then $\ln x^p = p\ln x$. Substitute $\ln A$ back in and we obtain the desired result.

(i) Wrong. (Put $A = 2$, $B = C = 1$.)

Question 26

Simplify the following expressions.

(a) $\exp [\ln(x)] - \ln[\exp(x)]$

(b) $\ln[x^4 \exp(-x)]$

(c) $\exp[\ln(x^2) - 2 \ln y]$

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(a) $\exp[\ln(x)] − \ln[\exp(x)] = e^{\ln x} − \ln e^x = x − x = 0$.

(b) $\ln[x^4 \exp(−x)] = 4 \ln x − x$.

(c) $\exp[\ln(x^2) - 2 \ln y] = \exp[\ln(x^2) - \ln(y^2)] = \exp[\ln(x^2/y^2)] = x^2/y^2$.