1. Since we have the same base on both sides of the equation \(\log _{117}(1-3x) = \log _{117}\left (x^2-3\right )\), we equate the arguments (what’s inside) of the logs to get \(1-3x = x^2-3\). Solving \(x^2+3x-4 = 0\) gives \(x=-4\) and \(x=1\).

   Using desmos, we graph both \(y = (x)\) and \(y = g(x)\) and find that these graphs intersect only at \(x=-4\).

   To see what happened to the solution \(x=1\), we substitute it into our original equation to obtain \(\log _{117}(-2) = \log _{117}(-2)\). While these expressions look identical, neither is a real number, ^[1] (They do, however, represent the same family of complex numbers. We refer the reader to a course in Complex Variables.) which means \(x=1\) is not in the domain of the original equation, and is not a solution.

2. To solve \(2 - \ln (t-3) = 1\), we first isolate the logarithm and get \(\ln (t-3) = 1\). Rewriting \(\ln (t-3) = 1\) as an exponential equation, we get is \(e^{1} = t-3\), so \(t =e+3\).

   Desmos shows the graphs of \(f(t) = 2 - \ln (t-3)\) and \(g(t) = 1\) intersect at \(t \approx 5.718 \approx e+3\).

3. We start solving \(\log _{6}(x+4) + \log _{6}(3-x) = 1\) by using the Product Rule for logarithms to rewrite the equation as \(\log _{6}\left [(x+4)(3-x)\right ] = 1\).

   Rewriting as an exponential equation gives \(6^{1} = (x+4)(3-x)\) which reduces to \(x^2+x-6 = 0\). We get two solutions: \(x=-3\) and \(x=2\).

   We graph \(y=f(x)\) and \(y=g(x)\) using desmos and see the graphs intersect twice, at \(x=-3\) and \(x=2\), as required.

4. Taking a cue from the previous problem, we begin solving \(\log _{7}(1-2t) = 1 - \log _{7}(3-t)\) by first collecting the logarithms on the same side, \(\log _{7}(1-2t) + \log _{7}(3-t) = 1\), and then using the Product Rule to get \(\log _{7}[(1-2t)(3-t)] = 1\).

   Rewriting as an exponential equation gives \(7^{1} = (1-2t)(3-t)\) which gives the quadratic equation \(2t^2-7t-4=0\). Solving, we find \(t = -\frac {1}{2}\) and \(t=4\).

   Once again, we use the change of base formula and find the graphs of \(y = f(t) = \frac {\ln (1-2t)}{\ln (7)}\) and \(y=g(t) = 1 - \frac {\ln (3-t)}{\ln (7)}\) intersect only at \(t=-\frac {1}{2}\).

   Checking \(t=4\) in the original equation produces \(\log _{7}(-7) = 1 - \log _{7}(-1)\), showing \(t=4\) is not in the domain of \(f\) nor \(g\).

5. Our first step in solving \(\log _{2}(x+3) = \log _{2}(6-x)+3\) is to gather the logarithms to one side of of the equation: \(\log _{2}(x+3) - \log _{2}(6-x) = 3\).

   The Quotient Rule gives \(\log _{2}\left (\frac {x+3}{6-x}\right ) = 3\) which, as an exponential equation is \(2^{3} = \frac {x+3}{6-x}\).

   Clearing denominators, we get \(8(6-x) = x+3\), which reduces to \(x = 5\).

   Graphing \(f\) and \(g\) using desmos, we find the graphs intersect at \(x=5\).

6. Our first step in solving \(1 + 2 \log _{4}(t+1) = 2 \log _{2}(t)\) is to gather the logs on one side of the equation. We obtain \(1 = 2 \log _{2}(t) - 2 \log _{4}(t+1)\) but find we need a common base to combine the logs.

   Since \(4\) is a power of \(2\), we use change of base to convert \(\log _{4}(t+1) = \frac {\log _{2}(t+1)}{\log _{2}(4)} = \frac {1}{2} \log _{2}(t+1)\). Hence, our original equation becomes

   \[ \begin{array}{rclr} 1 & = & 2 \log _{2}(t) - 2 \left (\frac {1}{2} \log _{2}(t+1)\right ) & \\ 1 &= & 2\log _{2}(t) - \log _{2}(t+1) & \\ 1 & = & \log _{2}\left (t^2\right ) - \log _{2}(t+1) & \text {Power Rule} \\ 1 & = & \log _{2}\left ( \dfrac {t^{2}}{t+1}\right ) & \text {Quotient Rule} \\ \end{array}\]

   Rewriting \(1 = \log _{2}\left ( \frac {t^{2}}{t+1}\right )\) in exponential form gives \( \frac {t^{2}}{t+1} = 2\) or \(t^2 -2t-2 = 0\). Using the quadratic formula, we obtain \(t = 1 \pm \sqrt {3}\).

   We use desmos to graph \(y = f(t)\) and \(y = g(t)\) below. We see the graphs intersect only at \(t \approx 2.732 \approx 1 + \sqrt {3}\).

   Note the solution \(t = 1 - \sqrt {3} < 0\) Hence if substituted into the original equation, the term \(2 \log _{2}\left (1 - \sqrt {3}\right )\) is undefined, which explains why the graphs intersect only once.

1. We start solving \(\frac {1}{\ln (x)+1} \leq 1\) by getting \(0\) on one side of the inequality: \(\frac {1}{\ln (x)+1} - 1 \leq 0\).

   Getting a common denominator yields \(\frac {1}{\ln (x)+1} - \frac {\ln (x)+1}{\ln (x)+1} \leq 0\) which reduces to \(\frac {-\ln (x)}{\ln (x)+1} \leq 0\), or \( \frac {\ln (x)}{\ln (x)+1} \geq 0\).

   We define \(r(x) = \frac {\ln (x)}{\ln (x)+1}\) and set about finding the domain and the zeros of \(r\). Due to the appearance of the term \(\ln (x)\), we require \(x > 0\). In order to keep the denominator away from zero, we solve \(\ln (x)+1 = 0\) so \(\ln (x) = -1\), so \(x = e^{-1} = \frac {1}{e}\). Hence, the domain of \(r\) is \(\left (0, \frac {1}{e}\right ) \cup \left (\frac {1}{e}, \infty \right )\).

   To find the zeros of \(r\), we set \(r(x) = \frac {\ln (x)}{\ln (x)+1} = 0\) so that \(\ln (x) = 0\), and we find \(x = e^{0} = 1\).

   In order to determine test values for \(r\) without resorting to the calculator, we need to find numbers between \(0\), \(\frac {1}{e}\), and \(1\) which have a base of \(e\). Since \(e \approx 2.718 > 1\), \(0 < \frac {1}{e^2} < \frac {1}{e} < \frac {1}{\sqrt {e}} < 1 < e\).

   To determine the sign of \(r\left ( \frac {1}{e^2} \right )\), note \(\ln \left (\frac {1}{e^2}\right ) = \ln \left (e^{-2}\right ) = -2\). Hence, \(r\left ( \frac {1}{e^2} \right ) = \frac {-2}{-2+1} = 2 > 0\). The rest of the test values are determined similarly.

   

   From our sign diagram, we find \(r(x) \geq 0\) on \(\left (0, \frac {1}{e}\right ) \cup [1, \infty )\), which is our solution.

   Graphing \(f(x) = \frac {1}{\ln (x)+1}\) and \(g(x) = 1\) using desmos, we see the graph of \(f\) is below the graph of \(g\) on these intervals, and that the graphs intersect at \(x=1\).

2. Moving all of the nonzero terms of \(\left (\log _{2}(x)\right )^2 < 2 \log _{2}(x) + 3\) to one side of the inequality in order to make use of a sign diagram, we have \(\left (\log _{2}(x)\right )^2 - 2 \log _{2}(x) - 3 < 0\).

   Defining \(r(x) = \left (\log _{2}(x)\right )^2 - 2 \log _{2}(x) - 3\), we get the domain of \(r\) is \((0, \infty )\), due to the presence of the logarithm. To find the zeros of \(r\), we set \(r(x) =\left (\log _{2}(x)\right )^2 - 2 \log _{2}(x) - 3= 0\) which we identify as a ‘quadratic in disguise.’

   Setting \(u = \log _{2}(x)\), our equation becomes \(u^2-2u-3 = 0\). Factoring gives us \(u=-1\) and \(u=3\). Since \(u = \log _{2}(x)\), we get \(\log _{2}(x) = -1\), or \(x = 2^{-1} = \frac {1}{2}\), and \(\log _{2}(x) = 3\), which gives \(x = 2^{3} = 8\).

   We use test values which are powers of \(2\): \(0 < \frac {1}{4} < \frac {1}{2} < 1 < 8 < 16\) to create the sign diagram below.

   

   From our sign diagram, we see \(r(x)< 0\), which corresponds to our solution, on \(\left (\frac {1}{2}, 8 \right )\).

   Using desmos, we graph \(f(x)= \left (\log _{2}(x)\right )^2\) and \(g(x) = 2 \log _{2}(x) + 3\). We find the graph of \(f\) is below the graph of \(g\) on \(\left (\frac {1}{2}, 8 \right )\).

3. We begin to solve \(t \log (t+1) \geq t\) by subtracting \(t\) from both sides to get \(t \log (t+1) - t \geq 0\).

   We define \(r(t) = t \log (t+1) - t \) and due to the presence of the logarithm, we require \(t > -1\).

   To find the zeros of \(r\), we set \(r(t) = t \log (t+1) - t = 0\). Factoring, we get \(t \left (\log (t+1) - 1\right ) = 0\), which gives \(t=0\) or \(\log (t+1) - 1=0\).

   From \(\log (t+1) - 1=0\) we get \(\log (t+1) = 1\), which we rewrite as \(t+1 = 10^{1}\). Hence, \(t = 9\).

   We select test values \(t\) so that \(t+1\) is a power of \(10\). Using \(-1 < -0.9 < 0 < \sqrt {10} -1 < 9 < 99\), we get the sign diagram below,

   

   Our sign diagram gives the solution as \((-1,0] \cup [9, \infty )\), which we check graphically using desmos.

   We find the graphs of \(y= f(t) = t \log (t+1)\) and \(y=g(t) = t\) intersect at \(t=0\) and \(t=9\) with the graph of \(f\) above the graph of \(g\) on the given solution intervals.

We require \(7.8 \leq -\log [\text {H}^{+}] \leq 8.5\) or \(-8.5 \leq \log [\text {H}^{+}] \leq -7.8\). One way to proceed is to break this compound inequality into two inequalities, solve each using a sign diagram, and take the intersection of the solution sets. ^[3] (Refer to page ?? for a discussion of what this means.)

On the other hand, we take advantage of the fact that \(F(x) = 10^{x}\) is an increasing function, meaning that if \(a \leq b \leq c\), then \(10^{a} \leq 10^{b} \leq 10^{c}\). This property allows us to solve our inequality in one step: from \(-8.5 \leq \log [\text {H}^{+}] \leq -7.8\), we get \(10^{-8.5} \leq 10^{\log [\text {H}^{+}]} \leq 10^{-7.8}\), so our solution is \( \leq [\text {H}^{+}] \leq 10^{-7.8}\). (Your Chemistry professor may want the answer written as \(3.16 \times 10^{-9} \leq [\text {H}^{+}] \leq 1.58 \times 10^{-8}\).) Using interval notation, our answer is \(\left [10^{-8.5}, 10^{-7.8}\right ]\).

After very carefully adjusting the viewing window on desmos, we see the graph of \(f(x) = -\log (x)\) lies between the lines \(y = 7.8\) and \(y = 8.5\) on the interval \([3.162 \times 10^{-9}, 1.5849 \times 10^{-8}]\).

1. We first write \(y=f(x)\) then interchange the \(x\) and \(y\) and solve for \(y\).

   \[ \begin{array}{rclr} y & = & f(x) & \\ y & = & \dfrac {\log (x)}{1-\log (x)} & \\ x & = & \dfrac {\log (y)}{1-\log (y)} & \text {Interchange $x$ and $y$.}\\ x\left (1-\log (y)\right ) & = & \log (y) & \\ x - x\log (y) & = & \log (y) & \\ x & = & x \log (y) + \log (y) & \\ x & = & (x+1) \log (y) & \\ \dfrac {x}{x+1} & = & \log (y) & \\ y & = & 10^{\frac {x}{x+1}} & \text {Rewrite as an exponential equation.}\\ \end{array}\]

   We have \(f^{-1}(x) = 10^{\frac {x}{x+1}}\). Graphing \(f\) and \(f^{-1}\) using desmos shows the required symmetry about the line \(y=x\).

2. Recognizing \(\frac {\log (x)}{1-\log (x)} = 1\) as \(f(x) = 1\), we have \(x = f^{-1}(1) = 10^{\frac {1}{1+1}} = 10^{\frac {1}{2}} = \sqrt {10}\).

   To check our answer algebraically, first recall \(\log (\sqrt {10}) = \log _{10}(\sqrt {10})\). Next, we know \(\sqrt {10} = 10^{\frac {1}{2}}\). Hence, \(\log _{10} \left (10^{\frac {1}{2}} \right ) = \frac {1}{2} = 0.5\). It follows that \(\frac {\log (\sqrt {10})}{1-\log (\sqrt {10})} = \frac {0.5}{1-0.5} = \frac {0.5}{0.5} = 1\), as required.

1. Owing to the presence of the ‘\(\ln (x)\)’ factor, we have the domain of \(f\) is \((0, \infty )\).
2. To find the \(x\)-intercepts, we set \(f(x) = x^2 \ln (x) = 0\). This gives \(x^2 = 0\) or \(\ln (x) = 0\), so \(x = 0\) or \(x = 1\). Since \(x=0\) is not in the domain of \(f\), our only solution is \(x = 1\). We have just one \(x\)-intercept, \((1,0)\).

3. To analyze \(\lim _{x \rightarrow 0^{+}} x^2 \ln (x)\), we note that as \(x \rightarrow 0^{+}\), \(x^2 \rightarrow 0\) but \(\ln (x) \rightarrow -\infty \). Hence we obtain the indeterminate form ^[4] (See the remarks following Example exponentialcurvesketchingex.) ‘\(0 \cdot (- \infty )\).’ Using desmos, we make a table of values of \(f(x)\) as \(x \rightarrow 0^{+}\). Both the table and the graph suggest that \(\lim _{x \rightarrow 0^{+}} x^2 \ln (x) = 0\). Since \(x = 0\) is not in the domain of \(f\), there is a hole in the graph at \((0,0)\).

4. To determine the intervals over which \(f\) is increasing and decreasing, we make a sign diagram for \(f'(x) = 2x \ln (x) + x\). Solving \(f'(x) = 2x \ln (x) + x = 0\) we factor: \(x(2\ln (x) + 1) = 0\). Since the domain of \(f\) is \((0, \infty )\), we focus on the factor \(2\ln (x) + 1 = 0\). We get \(\ln (x) = -\frac {1}{2}\) so \(x = e^{-\frac {1}{2}}\).

   When choosing test values, we could opt to find a decimal approximation for \(e^{-\frac {1}{2}}\) or we could use more ‘log-friendly’ values. In this case we note that \(e^{-1} < e^{-\frac {1}{2}} < e^{0} = 1\) so we choose \(e^{-1}\) and \(1\) as our test values.

   We find \(f'\left (e^{-1}\right ) = 2 \ln \left (e^{-1}\right ) + 1 = 2(-1) + 1 = -1 <0\) and \(f'(1) = 2 \ln (1) + 1 = 1 > 0\). Our sign diagram for \(f'(x)\) is below.

   

   We interpret the sign diagram using Theorem firstderivatveandgraphs as follows:

   

   We find \(f\) is decreasing on \(\left (0, e^{-\frac {1}{2}} \right )\) and increasing on \(\left ( e^{-\frac {1}{2}}, \infty \right )\).

5. We see \(f\) has a local (and in this case, global) minimum when \(x = e^{-\frac {1}{2}}\). The minimum value is \(f\left (e^{-\frac {1}{2}}\right ) = \left (e^{-\frac {1}{2}}\right )^2 \ln \left (e^{-\frac {1}{2}}\right ) = e^{-1} \left (-\frac {1}{2}\right ) = -\frac {e^{-1}}{2}\). The local minimum is \(\left ( e^{-\frac {1}{2}}, -\frac {e^{-1}}{2} \right )\).

6. To find the intervals over which the graph of \(f\) is concave up and concave down, we make a sign diagram for \(f''(x) = 2\ln (x) + 3\). Solving \(f''(x) = 2 \ln (x) + 3 = 0\) we get \(\ln (x) = -\frac {3}{2}\) so \(x = e^{-\frac {3}{2}}\). As with our first derivative analysis, we elect to choose some ‘log-friendly’ test values and note \(e^{-2} < e^{-\frac {3}{2}} < e^{0} = 1\).

   We find \(f''\left (e^{-2}\right ) = 2 \ln \left (e^{-2}\right ) + 3 = 2(-2) + 3 = -1 < 0\) and \(f''(1) = 2 \ln (1) + 3 = 3 > 0\).

   

   Interpreting our sign diagram for \(f''(x)\) using Theorem secondderivatveandgraphs gives:

   

   Hence, the graph of \(f\) is concave down on \(\left (0, e^{-\frac {3}{2}} \right )\) and concave up on \(\left (e^{-\frac {3}{2}} , \infty \right )\).

7. Since the concavity changes at \(x = e^{-\frac {3}{2}}\), we have an inflection point at \(\left ( e^{-\frac {3}{2}}, f \left (e^{-\frac {3}{2}}\right ) \right )\). We find \(f \left (e^{-\frac {3}{2}}\right ) = \left (f \left (e^{-\frac {3}{2}}\right ) \right )^2 \ln \left (f \left (e^{-\frac {3}{2}}\right ) \right ) = e^{-3} \left (-\frac {3}{2} \right ) = -\frac {3e^{-3}}{2}\). Our inflection point is \(\left ( e^{-\frac {3}{2}}, -\frac {3e^{-3}}{2} \right )\).

8. Using desmos, we confirm our calculations. ^[5] (It may require some window adjustment to capture the inflection point.)