Corrected exercises: exponential, inverse, logarithmic and inverse trigonometric functions (MATH 140)

Write every base as a power of $2$ BEFORE applying a law: $8^x = 2^{3x}$ and $4^{1-x} = 2^{2-2x}$. Then $\frac{2^{3x} \cdot 2^{2-2x}}{2^{x+3}} = 2^{3x + 2 - 2x - x - 3} = 2^{-1} = \frac{1}{2}$. The $x$ disappears completely: the expression is the constant $\frac{1}{2}$, which you can check at $x = 0$: $\frac{1 \cdot 4}{8} = \frac{1}{2}$. Next, $(e^{2x})^3 = e^{6x}$ and $\sqrt{e^{4x}} = (e^{4x})^{1/2} = e^{2x}$, so the quotient is $e^{6x - x - 2x} = e^{3x}$. Finally $\left(\frac{1}{9}\right)^{-x/2} = 9^{x/2} = (3^2)^{x/2} = 3^x$. The error to avoid is adding exponents of DIFFERENT bases: $8^x \cdot 4^{1-x}$ is not $32^{1}$, the laws only combine powers of the same base.

$3^{-x}$ is $3^x$ with $x$ replaced by $-x$: a reflection in the $y$-axis, giving the decreasing curve $\left(\frac{1}{3}\right)^x$. Then $-3^{-x}$ reflects that in the $x$-axis, and $2 - 3^{-x}$ shifts it up by $2$. The order matters for the last two: shifting first and reflecting second would give $-(3^{-x} + 2) = -2 - 3^{-x}$, another function. Domain: all reals, since $3^{-x}$ is defined everywhere. Range: $3^{-x}$ takes every value in $(0, \infty)$, so $-3^{-x}$ takes every value in $(-\infty, 0)$ and $g$ every value in $(-\infty, 2)$; the value $2$ itself is never reached, because $3^{-x}$ is never $0$. The $y$-intercept is $g(0) = 2 - 1 = 1$. As $x$ grows, $3^{-x}$ becomes as small as we like, so the graph approaches the horizontal line $y = 2$ from below without ever touching it. Check two points: $g(-1) = 2 - 3 = -1$ and $g(1) = 2 - \frac{1}{3} = \frac{5}{3}$, an increasing function, as the two reflections predict.

The two points give $C b = 6$ and $C b^3 = 24$. Divide the second equation by the first to eliminate $C$: $b^2 = 4$. The algebra offers $b = 2$ and $b = -2$; the base of an exponential function must be positive, so $b = -2$ is rejected, and $b = 2$. Then $C = \frac{6}{2} = 3$, so $f(x) = 3 \cdot 2^x$, and $f(-1) = \frac{3}{2}$. The figure agrees: the curve crosses the vertical axis at $f(0) = C = 3$. Two common errors: joining the points by a line, which gives slope $9$ and $f(-1) = -12$, a negative value that no exponential takes; and dividing the $y$-values as if the $x$-gap were $1$, which gives $b = 4$. The gap is $2$, so the ratio $4$ is $b^2$.

$(-2)^x$ is NOT an exponential function: $(-2)^{1/2} = \sqrt{-2}$ is not a real number, so it is not defined on an interval. $1^x = 1$ for every $x$ is a constant, excluded by the condition $b \ne 1$ (it would have no inverse, which is the point of the chapter). $\left(\frac{1}{2}\right)^x = 2^{-x}$ is exponential, with base in $(0, 1)$, hence decreasing. $x^2$ is a POWER function: the variable is in the base, not in the exponent. $\pi^x$ is exponential with base $\pi > 1$, increasing. $e^{-x} = \left(\frac{1}{e}\right)^x$ is exponential with base $\frac{1}{e} \in (0, 1)$. The test is always the same: where is the variable, and is the base a fixed positive number different from $1$?

(i) FALSE: $2^x + 2^x = 2 \cdot 2^x = 2^{x+1}$, while $4^x = 2^{2x}$; they agree only when $x + 1 = 2x$, that is at $x = 1$. At $x = 2$: $8 \ne 16$. There is no law for the SUM of two powers. (ii) FALSE: $(2^x)^3 = 2^{3x}$, the exponents multiply; at $x = 2$, $2^6 = 64$ while $2^{x^3} = 2^8 = 256$. (iii) TRUE: $3^x \cdot 3^x = 3^{2x} = (3^2)^x = 9^x$, or directly $(ab)^x = a^x b^x$ with $a = b = 3$. (iv) FALSE for every $x$: $e^{-x} = \frac{1}{e^x}$ is always positive, $-e^x$ always negative. A minus sign in the exponent means a RECIPROCAL, never a negative value; this is the error that makes students answer that $e^{-x}$ has negative values on its graph.

$f_1$: if $a^3 + 2 = b^3 + 2$ then $a^3 - b^3 = (a - b)(a^2 + ab + b^2) = 0$. The second factor equals $\left(a + \frac{b}{2}\right)^2 + \frac{3b^2}{4}$, which is $0$ only when $a = b = 0$; so in every case $a = b$, and $f_1$ is one-to-one. $f_2(x) = x(x - 4)$ gives $f_2(0) = f_2(4) = 0$: two inputs, one output, NOT one-to-one; the horizontal line $y = 0$ meets its parabola twice. $f_3(0) = f_3(2) = 1$: not one-to-one. $f_4$: $\frac{1}{a} = \frac{1}{b}$ gives $a = b$ directly, so $f_4$ is one-to-one although its graph is in two pieces. One counterexample settles a no; a yes needs the algebra for ALL inputs, and a few matching values prove nothing.

Write $y = \frac{2x + 1}{x - 3}$ and solve for $x$: $y(x - 3) = 2x + 1$, so $xy - 2x = 3y + 1$, $x(y - 2) = 3y + 1$ and $x = \frac{3y + 1}{y - 2}$, possible only when $y \ne 2$. Renaming, $f^{-1}(x) = \frac{3x + 1}{x - 2}$. Its domain is $x \ne 2$, which is the range of $f$ (the value $2$ is never taken: $\frac{2x + 1}{x - 3} = 2$ would give $1 = -6$). Its range is $y \ne 3$, the domain of $f$. Check: $f(4) = \frac{9}{1} = 9$ and $f^{-1}(9) = \frac{28}{7} = 4$. The step students skip is collecting the $x$ terms on ONE side before factoring; without it they end with $x$ on both sides.

Complete the square: $x^2 - 4x = (x - 2)^2 - 4$. On $[2, \infty)$ the function increases from $-4$, so its range is $[-4, \infty)$. Solve $y = (x - 2)^2 - 4$: $(x - 2)^2 = y + 4$, so $x - 2 = \pm\sqrt{y + 4}$. Here the domain decides the sign: $x \ge 2$ means $x - 2 \ge 0$, so $x = 2 + \sqrt{y + 4}$, and the inverse is $x \mapsto 2 + \sqrt{x + 4}$ on $[-4, \infty)$. On $(-\infty, 2]$, $x - 2 \le 0$ and the other sign is forced: $2 - \sqrt{x + 4}$. Checks: $f_2(5) = 5$ and $2 + \sqrt 9 = 5$; $f_2(-1) = 5$ and $2 - \sqrt 9 = -1$. Writing $2 \pm \sqrt{x + 4}$ is not an answer: it gives two outputs for one input, so it is not a function.

Read the figure backwards: $f^{-1}(2)$ is the input whose output is $2$, and the marked point $(1, 2)$ gives $f^{-1}(2) = 1$. Likewise $(-3, 0)$ gives $f^{-1}(0) = -3$. The range of $f$ on $[-3, 6]$ is $[0, 3]$, from $f(-3) = 0$ to $f(6) = 3$, so $f^{-1}$ has domain $[0, 3]$ and range $[-3, 6]$. Formula: $y = \sqrt{x + 3}$ gives $y^2 = x + 3$ with $y \ge 0$, so $f^{-1}(x) = x^2 - 3$ for $0 \le x \le 3$. The restriction is part of the answer: $x^2 - 3$ on all of $\mathbb{R}$ is a parabola that is not one-to-one, and it is not the inverse of anything.

Each point $(a, b)$ of $f$ gives the point $(b, a)$ of $f^{-1}$: $(0, -3)$, $(1, -2)$, $(2, 1)$, $(3, 6)$, and the graph of $f^{-1}$ is the mirror image of $f$ in the line $y = x$ (figure of the solution, same scale on both axes, otherwise the mirror is distorted). For $x \in [-3, 6]$: $f^{-1}(f(x)) = (\sqrt{x + 3})^2 - 3 = x$. For $x \in [0, 3]$: $f(f^{-1}(x)) = \sqrt{x^2 - 3 + 3} = \sqrt{x^2} = |x| = x$ because $x \ge 0$. Outside $[0, 3]$ the formula would give $\sqrt{x^2} = |x|$, which is $-x$ for negative $x$: at $x = -2$, $\sqrt{(-2)^2} = 2$. The cancellation equations hold only on the right domains, and that is the thread of the whole chapter.

Each value is a question about an exponent. $\log_2 \frac{1}{32}$: $2^y = 2^{-5}$, so $-5$. $\log_9 27$: $9^y = 27$ means $3^{2y} = 3^3$, so $y = \frac{3}{2}$. $\ln \frac{1}{\sqrt e} = \ln e^{-1/2} = -\frac{1}{2}$. $e^{3\ln 2} = (e^{\ln 2})^3 = 2^3 = 8$, by the cancellation $e^{\ln a} = a$ for $a > 0$. $\log_{10} 0.001 = \log_{10} 10^{-3} = -3$. A negative logarithm is normal: it says the number is between $0$ and $1$. What does not exist is the logarithm OF a negative number.

For $x > 2$ every factor is positive, and the laws apply one by one: $\ln x^3 + \ln \sqrt{x + 1} - \ln e^2 - \ln(x - 2) = 3\ln x + \frac{1}{2}\ln(x + 1) - 2 - \ln(x - 2)$. The condition matters, because the left side has a larger domain than the right. For $-1 < x < 0$ the numerator $x^3\sqrt{x + 1}$ and the denominator $e^2(x - 2)$ are both negative, so the quotient is positive: at $x = -\frac{1}{2}$ it equals $\frac{(-1/8)\sqrt{1/2}}{e^2(-5/2)} > 0$, the left side exists, and $\ln x$ on the right does not. The expansion is therefore valid for $x > 2$ only, where every piece exists; announcing that condition is worth the mark here.

$\log_2 40 - \log_2 5 = \log_2 8 = 3$. $\ln 12 - 2\ln 2 + \ln \frac{1}{3} = \ln \frac{12}{4 \cdot 3} = \ln 1 = 0$. For the last one, turn every term into a logarithm first, including the constant: $1 = \ln e$. Then $3 \ln x - \frac{1}{2}\ln(x^2 + 1) + \ln e = \ln \frac{e x^3}{\sqrt{x^2 + 1}}$, valid for $x > 0$ (the original needs $\ln x$). Forgetting that $1 = \ln e$ and writing $\ln \frac{x^3}{\sqrt{x^2+1}} + 1$ is not wrong, but it is not a single logarithm, and the question loses its mark.

$\ln(x^2 - 4x)$ needs $x(x - 4) > 0$: both factors positive ($x > 4$) or both negative ($x < 0$). Its domain is $(-\infty, 0) \cup (4, \infty)$: the two branches of the figure. The student's $\ln x + \ln(x - 4)$ needs $x > 0$ AND $x > 4$, that is $x > 4$ only. The two expressions agree on $(4, \infty)$, but the student's formula has thrown away the whole left branch: at $x = -1$, $\ln(1 + 4) = \ln 5$ exists while $\ln(-1)$ does not. The law $\ln(ab) = \ln a + \ln b$ needs $a > 0$ AND $b > 0$; when only the product is positive, write $\ln(ab) = \ln|a| + \ln|b|$, here $\ln|x| + \ln|x - 4|$, valid on the whole domain.

By change of base, $\log_3 5 \cdot \log_5 9 = \frac{\ln 5}{\ln 3} \cdot \frac{\ln 9}{\ln 5} = \frac{\ln 9}{\ln 3} = \frac{2\ln 3}{\ln 3} = 2$. The long product telescopes the same way: $\frac{\ln 3}{\ln 2} \cdot \frac{\ln 4}{\ln 3} \cdots \frac{\ln 32}{\ln 31} = \frac{\ln 32}{\ln 2} = \log_2 32 = 5$. Finally, $\log_2$ is increasing and $4 < 7 < 8$, so $\log_2 4 < \log_2 7 < \log_2 8$, that is $2 < \log_2 7 < 3$. The exact value is $\frac{\ln 7}{\ln 2}$, and bracketing it by powers of the base is the no-calculator way to know its size.