Exponents and Logarithms
Solutions

1. We have ${\log }_x(2\cdot 4\cdot 8)=1$ which implies that ${\log }_x(64)=1$. Therefore, $x=64$.

2. Since $12=2^2\cdot 3$ it follows that $$\begin{array}{rl}{12}^{2x+1}& =2^{3x+7}\cdot 3^{3x-4}\\ 2^{2(2x+1)}\cdot 3^{2x+1}& =2^{3x+7}\cdot 3^{3x-4}\\ 2^{2(2x+1)-3x-7}& =3^{3x-4-2x-1}\\ 2^{x-5}& =3^{x-5}\end{array}$$ The graphs of $y=2^{x-5}$ and $y=3^{x-5}$ intersect only at $x=5$ and $y=1$. (Since $2^z=3^z$ only for $z=0$.) Therefore, $x=5$ is the only solution.

3. This expression equals ${\log }_{10}({\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{4}{3}}\cdot {\displaystyle \frac{5}{4}}\cdot \cdots \cdot {\displaystyle \frac{200}{199}})={\log }_{10}{\displaystyle \left({\displaystyle \frac{200}{2}}\right)={\log }_{10}100=2}$.

4. For the first equation we know that $x,y\ne 0$. The second equation states $x{y}^{-2}=3^{-3}$ which gives us $x^3{y}^{-6}=3^{-9}$ after cubing both sides. Since neither side of this new equation is $0$ we can divide the first equation by this new equation to eliminate $x$. $$\begin{array}{rl}{\displaystyle \frac{x^3{y}^5}{x^3{y}^{-6}}}& ={\displaystyle \frac{2^{11}{3}^{13}}{3^{-9}}}\\ y^{11}& =2^{11}\cdot 3^{22}\\ y^{11}& =2^{11}\cdot (3^2{)}^{11}\\ y^{11}& =(2\cdot 9{)}^{11}\\ y& =18.\end{array}$$ Therefore, $x={\displaystyle \frac{y^2}{27}}=12$.

5. ${\log }_8(18)={\log }_{8}2+{\log }_{8}9={\displaystyle \frac{1}{3}}+{\log }_8{3}^2={\displaystyle \frac{1}{3}}+2{\log }_{8}3={\displaystyle \frac{1}{3}}+2k=2k+{\displaystyle \frac{1}{3}}$

6. Solution 1

   We express the logarithms in exponential form to arrive at: $2x=2^y$ and $x=4^y$. Thus, $$\begin{array}{rl}{2}^y& =2(4^y)\\ 2^y& =2(2^{2y})\\ 2^y& =2^{2y+1}\\ y& =2y+1\\ y& =-1\end{array}$$ Thus, $x=4^{-1}={\displaystyle \frac{1}{4}}$. Therefore, the point of intersection is $({\displaystyle \frac{1}{4}},-1)$.

   Solution 2

   Substituting one equation into the other we obtain $$\begin{array}{rl}{\log }_{2}2x& ={\log }_{4}x\\ {\displaystyle \frac{\log 2x}{\log 2}}& ={\displaystyle \frac{\log x}{\log 4}}\\ {\displaystyle \frac{\log 2x}{\log 2}}& ={\displaystyle \frac{\log x}{\log 2^2}}\\ {\displaystyle \frac{\log 2x}{\log 2}}& ={\displaystyle \frac{\log x}{2\log 2}}\\ 2\log 2x& =\log x\\ 2\log 2+2\log x& =\log x\\ \log x& =-2\log 2\\ \log x& =\log 2^{-2}\\ \log x& =\log {\displaystyle \frac{1}{4}}\end{array}$$ Therefore, $x={\displaystyle \frac{1}{4}}$. Substituting this value back into either of the original equations and we obtain that $y=-1$. Therefore, the point of intersection is $({\displaystyle \frac{1}{4}},-1)$.

7. We note first that $x=a^y$ for all points on the curve. The midpoint of $AB$ is given by $({\displaystyle \frac{x_1+x_2}{2}}$,${\displaystyle \frac{y_1+y_2}{2}})$. Since we draw a horizontal line through the midpoint, $y_3={\displaystyle \frac{y_1+y_2}{2}}$, we have that $$\begin{array}{rl}(x_3{)}^2& =(a^{y_3}{)}^2\\ & =(a^{\frac{y_1+y_2}{2}}{)}^2\\ & =(a^{y_1})(a^{y_2})\\ & =x_1{x}_2.\end{array}$$

8. Given that the graph of the function passes through $(2,1)$, we know that $a\ne 0$. We have $1=a(2^r)$ and $4=a({32}^r)$. Since neither side of the first equation is 0, we can divide the second equation by the first to obtain $4={\displaystyle \frac{{32}^r}{2^r}}={\displaystyle \frac{2^r\cdot {16}^r}{2^r}}={16}^r$. Therefore, $r={\displaystyle \frac{1}{2}}$.

9. Factoring both sides of the equation $2^{x+3}+2^x=3^{y+2}-3^y$ gives $$\begin{array}{rl}(2^3+1)2^x& =(3^2-1)3^y\\ 9\cdot 2^x& =8\cdot 3^y\\ 3^2\cdot 2^x& =2^3\cdot 3^y\\ 2^{x-3}& =3^{y-2}\end{array}$$ Since $x$ and $y$ are integers, and the only integer power of 2 that is also an integer power of 3 is the number $1=2^0=3^0$, we have $x=3$ and $y=2$.

10. If $f(x)=2^{4x-2}$, then $f(x)\cdot f(1-x)=2^{4x-2}\cdot 2^{4(1-x)-2}=2^{4x-2+4-4x-2}=2^0=1$.

11. Observe that the argument of both logarithms must be positive and so $x>6$. Now $$\begin{array}{rl}{\log }_5(x-2)+{\log }_5(x-6)& =2\\ {\log }_5((x-2)(x-6))& =2\\ (x-2)(x-6)& =25\\ x^2-8x-13& =0\\ x& =4\pm \sqrt{29}\end{array}$$ However, since $x>6$, we have that $x=4+\sqrt{29}$.

12. If $a,b,c$ is a geometric sequence, then ${\displaystyle \frac{b}{a}}={\displaystyle \frac{c}{b}}$. It follows that ${\log }_x({\displaystyle \frac{b}{a}})={\log }_x({\displaystyle \frac{c}{b}})$ which implies ${\log }_{x}b-{\log }_{x}a={\log }_{x}c-{\log }_{x}b$. Therefore, the logarithms form an arithmetic sequence.

    If ${\log }_{x}a,{\log }_{x}b,{\log }_{x}c$ form an arithmetic sequence, then $$\begin{array}{rl}{\log }_{x}b-{\log }_{x}a& ={\log }_{x}c-{\log }_{x}b\\ {\log }_x\left({\displaystyle \frac{b}{a}}\right)& ={\log }_x\left({\displaystyle \frac{c}{b}}\right)\\ {\displaystyle \frac{b}{a}}& ={\displaystyle \frac{c}{b}}& \text{(since the log function takes on each value only once)}\end{array}$$ Thus, $a,b,c$ form a geometric sequence.

13. Using exponent rules and arithmetic, we manipulate the given equation: $$\begin{array}{rl}{3}^{x+2}+2^{x+2}+2^x& =2^{x+5}+3^x\\ 3^x{3}^2+2^x{2}^2+2^x& =2^x{2}^5+3^x\\ 9(3^x)+4(2^x)+2^x& =32(2^x)+3^x\\ 8(3^x)& =27(2^x)\\ {\displaystyle \frac{3^x}{2^x}}& ={\displaystyle \frac{27}{8}}\\ {\left({\displaystyle \frac{3}{2}}\right)}^x& ={\left({\displaystyle \frac{3}{2}}\right)}^3\end{array}$$ Since the two expressions are equal and the bases are equal, then the exponents must be equal, and so $x=3$.

14. Let $a={\log }_{10}x$. Then $({\log }_{10}x{)}^{{\log }_{10}({\log }_{10}x)}=10000$ becomes $a^{{\log }_{10}a}={10}^4$.

    Taking the base $10$ logarithm of both sides and using the fact that ${\log }_{10}(a^b)=b{\log }_{10}a$, we obtain $({\log }_{10}a)({\log }_{10}a)=4$ or $({\log }_{10}a{)}^2=4$. Therefore, ${\log }_{10}a=\pm 2$ and so ${\log }_{10}({\log }_{10}x)=\pm 2$.

    If ${\log }_{10}({\log }_{10}x)=2$, then ${\log }_{10}x={10}^2=100$ and so $x={10}^{100}$.

    If ${\log }_{10}({\log }_{10}x)=-2$, then ${\log }_{10}x={10}^{-2}=\frac{1}{100}$ and so $x={10}^{1/100}$. Therefore, $x={10}^{100}$ or $x={10}^{1/100}$.

15. Note that $x\ne 1$ since $1$ cannot be the base of a logarithm. This tells us that $\log x\ne 0$.

    Using the fact that ${\log }_{a}b={\displaystyle \frac{\log b}{\log a}}$ and then using other logarithm laws, we obtain the following equivalent equations: $$\begin{array}{rl}{\log }_{4}x-{\log }_{x}16& ={\displaystyle \frac{7}{6}}-{\log }_{x}8\\ {\displaystyle \frac{\log x}{\log 4}}-{\displaystyle \frac{\log 16}{\log x}}& ={\displaystyle \frac{7}{6}}-{\displaystyle \frac{\log 8}{\log x}}& \text{(note that }x\ne 1\text{, so }\log x\ne 0\text{)}\\ {\displaystyle \frac{\log x}{\log 4}}& ={\displaystyle \frac{7}{6}}+{\displaystyle \frac{\log 16-\log 8}{\log x}}\\ {\displaystyle \frac{\log x}{\log (2^2)}}& ={\displaystyle \frac{7}{6}}+{\displaystyle \frac{\log (\frac{16}{8})}{\log x}}\\ {\displaystyle \frac{\log x}{2\log 2}}& ={\displaystyle \frac{7}{6}}+{\displaystyle \frac{\log 2}{\log x}}\\ {\displaystyle \frac{1}{2}}\left({\displaystyle \frac{\log x}{\log 2}}\right)& ={\displaystyle \frac{7}{6}}+{\displaystyle \frac{\log 2}{\log x}}\end{array}$$ Letting $t={\displaystyle \frac{\log x}{\log 2}}={\log }_{2}x$ and noting that $t\ne 0$ since $x\ne 1$, we obtain the following equations equivalent to the previous ones: $$\begin{array}{rl}{\displaystyle \frac{t}{2}}& ={\displaystyle \frac{7}{6}}+{\displaystyle \frac{1}{t}}\\ 3t^2& =7t+6& \text{(multiplying both sides by }6t\text{)}\\ 3t^2-7t-6& =0\\ (3t+2)(t-3)& =0\end{array}$$ Therefore, the original equation is equivalent to $t=-\frac{2}{3}$ or $t=3$.

    Converting back to the variable $x$, we obtain ${\log }_{2}x=-\frac{2}{3}$ or ${\log }_{2}x=3$, which gives $x=2^{-2/3}$ or $x=2^3=8$.