Solution. Every one of these is the basic curve \(y=a^x\) moved, flipped or both. The basic curve passes through \((0,1)\), is positive everywhere, and has the \(x\)-axis as a horizontal asymptote. Whether it rises or falls depends on the base: greater than \(1\) and it rises, between \(0\) and \(1\) and it falls.

Two worth working through.

(iii). \(5^{-x}=\left (5^{-1}\right )^x=\left (\frac {1}{5}\right )^x\). A negative index in the exponent is the same as reflecting in the \(y\)-axis, so this is \(5^x\) read backwards — falling instead of rising, still through \((0,1)\).

(viii). \(f(x)=2^{x-1}-3\). The \(x-1\) shifts the curve one unit to the right, and the \(-3\) shifts it three units down, taking the asymptote with it to \(y=-3\). The \(y\)-intercept is \(f(0)=2^{-1}-3=\frac {1}{2}-3=-\frac {5}{2}\), and the curve crosses the \(x\)-axis where \(2^{x-1}=3\), that is \(x=1+\log _2 3\approx 2.58\).

Solution. The method is to write both sides as powers of the same base and then equate the indices.

(i). \(256=4^4\), so \[4^x=4^4\quad \implies \quad x=4.\]

(ii). This one is different — the unknown is in the base, not the index, so it is not an exponential equation at all. Take cube roots: \[x=\sqrt [3]{125}=5.\]

(iii). \(\frac {1}{16}=\left (\frac {1}{2}\right )^4\), so \[\left (\tfrac {1}{2}\right )^x=\left (\tfrac {1}{2}\right )^4\quad \implies \quad x=4.\]

(iv). \(0.125=\frac {1}{8}=2^{-3}\), so \[2^x=2^{-3}\quad \implies \quad x=-3.\]

(v). Write everything as a power of \(3\), using \(27=3^3\) and \(9=3^2\): \[\left (3^3\right )^{4x}=\left (3^2\right )^{x+1}\quad \implies \quad 3^{12x}=3^{2x+2},\] \[12x=2x+2\quad \implies \quad 10x=2\quad \implies \quad x=\tfrac {1}{5}.\]

(vi). Write everything as a power of \(2\), using \(\frac {1}{8}=2^{-3}\): \[\left (2^{-3}\right )^{-2t}=2^{t+3}\quad \implies \quad 2^{6t}=2^{t+3},\] \[6t=t+3\quad \implies \quad 5t=3\quad \implies \quad t=\tfrac {3}{5}.\]

Solution. Each of these is a quadratic in disguise. The key identities are \[2^{2x}=\left (2^x\right )^2,\qquad 4^x=\left (2^2\right )^x=\left (2^x\right )^2,\qquad 2^{x+1}=2\cdot 2^x.\] So put \(u=2^x\) throughout, remembering that \(u>0\) always — a power of \(2\) is never zero or negative.

(i). With \(u=2^x\): \[u^2+3u-4=0\quad \implies \quad (u+4)(u-1)=0\quad \implies \quad u=-4 \text { or } u=1.\] Reject \(u=-4\), since \(2^x>0\). Then \[2^x=1\quad \implies \quad x=0.\]

(ii). Rewrite: \(4^x=u^2\) and \(2^{x+1}=2u\): \[u^2-2u=48\quad \implies \quad u^2-2u-48=0\quad \implies \quad (u-8)(u+6)=0.\] Reject \(u=-6\); then \(u=8\), so \[2^x=8=2^3\quad \implies \quad x=3.\]

(iii). \[u^2-6u-16=0\quad \implies \quad (u-8)(u+2)=0.\] Reject \(u=-2\); then \(2^x=8\), so \[x=3.\]

(iv). Here \(2^{2x+1}=2\cdot 2^{2x}=2u^2\) and \(2^{x+1}=2u\): \[2u^2-2u+1=u\quad \implies \quad 2u^2-3u+1=0\quad \implies \quad (2u-1)(u-1)=0,\] \[u=\tfrac {1}{2}\quad \text {or}\quad u=1.\] Both are positive, so both survive: \[2^x=\tfrac {1}{2}=2^{-1}\ \implies \ x=-1,\qquad 2^x=1\ \implies \ x=0.\] \[\therefore \quad x=-1 \text { or } x=0.\]

Solution. The two forms say the same thing: \[a^b=c\qquad \Longleftrightarrow \qquad \log _a c=b.\] In words: the logarithm is the index. The base stays the base, and the answer to the logarithm is the power.

(a). \[\text {(i) } y=5^{x+2}\quad \implies \quad \log _5 y=x+2\] \[\text {(ii) } \left (\tfrac {3}{4}\right )^{-2}=\tfrac {16}{9}\quad \implies \quad \log _{\frac {3}{4}}\left (\tfrac {16}{9}\right )=-2\] \[\text {(iii) } y=e^x\quad \implies \quad \log _e y=x,\ \text { usually written } \ln y=x\] \[\text {(iv) } 8^{-1}=0.125\quad \implies \quad \log _8(0.125)=-1\]

(b). \[\text {(i) } \log _2 32=5\quad \implies \quad 2^5=32\] \[\text {(ii) } \log _2\left (\tfrac {1}{16}\right )=-4\quad \implies \quad 2^{-4}=\tfrac {1}{16}\] \[\text {(iii) } \log _{27}3=\tfrac {1}{3}\quad \implies \quad 27^{\frac {1}{3}}=3\]

(c). Each asks: to what power must the base be raised? \[\text {(i) } 3^2=9\quad \implies \quad \log _3 9=2\] \[\text {(ii) } 5^3=125\quad \implies \quad \log _5 125=3\] \[\text {(iii) } 6^{-2}=\tfrac {1}{36}\quad \implies \quad \log _6\left (\tfrac {1}{36}\right )=-2\]

Solution. Convert to exponential form and evaluate.

\[\text {(i) } \log _3 x=4\quad \implies \quad x=3^4=81.\]

\[\text {(ii) } \log _4 x=\tfrac {3}{2}\quad \implies \quad x=4^{\frac {3}{2}}=\left (\sqrt {4}\right )^3=2^3=8.\]

(iii). Here the unknown is the base: \[\log _a 2=\tfrac {1}{2}\quad \implies \quad a^{\frac {1}{2}}=2\quad \implies \quad \sqrt {a}=2\quad \implies \quad a=4.\]

\[\text {(iv) } \log _2(3x+4)=4\quad \implies \quad 3x+4=2^4=16\quad \implies \quad 3x=12\quad \implies \quad x=4.\] Check the argument is positive: \(3(4)+4=16>0\). \(\checkmark \)

\[\text {(v) } \ln x=-2\quad \implies \quad x=e^{-2}\approx 0.135.\]

Solution. The three laws needed throughout are \[\log _a M+\log _a N=\log _a(MN),\qquad \log _a M-\log _a N=\log _a\!\left (\frac {M}{N}\right ), \qquad \log _a\left (M^k\right )=k\log _a M.\]

(i). Combine the left side into one logarithm: \[\log _5\left (\frac {x}{4}\right )=2\quad \implies \quad \frac {x}{4}=5^2=25 \quad \implies \quad x=100.\]

(ii). The unknown is the base. Combining, \[\log _x(8\times 4)=\log _x 32=5\quad \implies \quad x^5=32\quad \implies \quad x=2.\]

(iii). Write the \(1\) as a logarithm to the same base: \(1=\log _3 3\). Then \[\log _3\left (x^2+2\right )=\log _3 3+\log _3(x+2)=\log _3\left [3(x+2)\right ].\] Since the logarithms have the same base, their arguments are equal: \[x^2+2=3x+6\quad \implies \quad x^2-3x-4=0\quad \implies \quad (x-4)(x+1)=0,\] giving \(x=4\) or \(x=-1\).

Check both against the domain. Every argument must be positive. At \(x=4\): \(x^2+2=18>0\) and \(x+2=6>0\). \(\checkmark \) At \(x=-1\): \(x^2+2=3>0\) and \(x+2=1>0\). \(\checkmark \) \[\therefore \quad x=4\ \text { or }\ x=-1.\]

(iv). Here \(2=\log _4 16\), so \[\log _4\left (x^2+8x-1\right )=\log _4\left [16(x-1)\right ],\] \[x^2+8x-1=16x-16\quad \implies \quad x^2-8x+15=0\quad \implies \quad (x-3)(x-5)=0,\] giving \(x=3\) or \(x=5\).

Check. At \(x=3\): \(x-1=2>0\) and \(x^2+8x-1=32>0\). \(\checkmark \) At \(x=5\): \(x-1=4>0\) and \(x^2+8x-1=64>0\). \(\checkmark \) \[\therefore \quad x=3\ \text { or }\ x=5.\]

(v). The second logarithm has base \(\frac {1}{3}\). Change it to base \(3\): \[\log _{\frac {1}{3}}x^2=\frac {\log _3 x^2}{\log _3\frac {1}{3}}=\frac {2\log _3 x}{-1} =-2\log _3 x,\] since \(\log _3\frac {1}{3}=\log _3 3^{-1}=-1\). Substituting, \[\log _3 x-\left (-2\log _3 x\right )=3\log _3 x=6\quad \implies \quad \log _3 x=2 \quad \implies \quad x=9.\]

(vi). Change every term to base \(2\), using \(4=2^2\) and \(16=2^4\): \[\log _4 x=\frac {\log _2 x}{\log _2 4}=\frac {\log _2 x}{2},\qquad \log _{16}x=\frac {\log _2 x}{\log _2 16}=\frac {\log _2 x}{4}.\] Writing \(L=\log _2 x\), \[L+\frac {L}{2}+\frac {L}{4}=21\quad \implies \quad \frac {4L+2L+L}{4}=\frac {7L}{4}=21 \quad \implies \quad L=12,\] \[\therefore \quad x=2^{12}=4096.\]

Solution. (a). Every logarithmic graph passes through \((1,0)\), has the \(y\)-axis as a vertical asymptote, and is defined only for \(x>0\). A base greater than \(1\) gives a rising curve; a base between \(0\) and \(1\) gives a falling one.

(i). \(\log _7 x\): rising, through \((1,0)\) and \((7,1)\).

(ii) and (iii) are the same curve. By change of base, \[\log _{\frac {1}{7}}x=\frac {\log _7 x}{\log _7\frac {1}{7}}=\frac {\log _7 x}{-1} =-\log _7 x,\] so (ii) is (i) reflected in the \(x\)-axis. And (iii), \(-\log _3 x\), is the same reflection applied to base \(3\) — which by the same identity equals \(\log _{\frac {1}{3}}x\), that is (iv).

\[\therefore \quad \text {(iii) and (iv) are identical functions.}\]

(b). In each of (i) and (ii), \(f\) and \(g\) are inverses of one another, so their graphs are reflections in the line \(y=x\). Where \(f\) passes through \((0,1)\), \(g\) passes through \((1,0)\); where \(f\) has the horizontal asymptote \(y=0\), \(g\) has the vertical asymptote \(x=0\).

(iii) is the exception. \(e^x\) and \(\ln x\) are inverses, but \(\ln (x-3)\) is \(\ln x\) shifted three units to the right, with vertical asymptote \(x=3\) and \(x\)-intercept at \(x=4\). So \(g\) is not the inverse of \(f\) here, and the two graphs are not reflections of each other in \(y=x\).

Solution. (a). Set the model equal to \(25\,000\): \[12500e^{0.015t}=25000\quad \implies \quad e^{0.015t}=2.\] Take natural logarithms of both sides, which is what undoes \(e\): \[0.015t=\ln 2\quad \implies \quad t=\frac {\ln 2}{0.015}=\frac {0.6931}{0.015} \approx 46.2.\] \[\therefore \quad \text {the population reaches } 25\,000 \text { after about } 46 \text { years}.\] Note this is the time to double, and it does not depend on the starting figure of \(12\,500\) — the \(12500\) cancelled at the first step.

(b). Here the growth rate is not given and must be found first. Write \[p(t)=20e^{kt}.\]

Find \(k\) from the information at \(t=100\): \[20e^{100k}=80\quad \implies \quad e^{100k}=4\quad \implies \quad 100k=\ln 4\] \[\implies \quad k=\frac {\ln 4}{100}=\frac {1.3863}{100}\approx 0.013863.\]

Now find \(t\) when \(p=200\): \[20e^{kt}=200\quad \implies \quad e^{kt}=10\quad \implies \quad kt=\ln 10\] \[\implies \quad t=\frac {\ln 10}{k}=\frac {2.3026}{0.013863}\approx 166.1.\] \[\therefore \quad \text {the population reaches } 200 \text { after about } 166 \text { days}.\]

(c). Substitute \(W_0=25\) and \(t=4\): \[W=25e^{-0.25(4)}=25e^{-1}=\frac {25}{2.7183}\approx 9.20.\] \[\therefore \quad \text {the wound is about } 9.2\text { mm}^2 \text { after } 4 \text { days}.\]