Exponents and Logarithms

Toolkit

Exponents

Let $a$, $b$, $x$, and $y$ be real numbers and let $n$ be an integer with $n\ge 2$. The rules for exponents are

• $a^{\frac{1}{n}}=\sqrt[n]{a}$

• $a^0=1$, if $a\ne 0$

• $a^{-x}={\displaystyle \frac{1}{a^x}}$, if $a\ne 0$

• $a^x{a}^y=a^{x+y}$

• ${\displaystyle \frac{a^x}{a^y}}=a^{x-y}$, if $a\ne 0$

• $(a^x{)}^y=a^{xy}$

• $a^x\cdot b^x=(ab{)}^x$

• ${\displaystyle \frac{a^x}{b^x}}={\left({\displaystyle \frac{a}{b}}\right)}^x$, if $b\ne 0$

Also, $0^0$ is not defined, if it is encountered using any of the above formulae.

Logarithms

Let $x$ and $y$ be positive real numbers. Let $a$ be a positive real number with $a\ne 1$. The equation $y=a^x$ is equivalent to ${\log }_{a}y=x$. The rules for logarithms are

• ${\log }_{a}1=0$

• ${\log }_{a}a=1$

• ${\log }_a(xy)={\log }_{a}x+{\log }_{a}y$

• ${\log }_a\left({\displaystyle \frac{x}{y}}\right)={\log }_{a}x-{\log }_{a}y$

• ${\log }_a(x^y)=y{\log }_{a}x$

• ${\log }_a(a^x)=x$

• $a^{{\log }_{a}x}=x$

• ${\log }_{a}x={\displaystyle \frac{1}{{\log }_{x}a}}$, where $x\ne 1$

• ${\log }_{y}x={\displaystyle \frac{{\log }_{a}x}{{\log }_{a}y}}$

Sample Problems

1. Given $2{\log }_5(x-3y)={\log }_5(2x)+{\log }_5(2y)$, calculate the ratio ${\displaystyle \frac{x}{y}}$.

   Solution

   First, we note that in the original equation, if the three logarithmic terms are to be defined, then their arguments must be positive. So $x>0$, $y>0$, and $x>3y$. Now $$\begin{array}{rl}2{\log }_5(x-3y)& ={\log }_5(2x)+{\log }_5(2y)\\ {\log }_5(x-3y{)}^2& ={\log }_5(4xy).\end{array}$$

   Now since the log function takes on each value in its range only once, we have that $$\begin{array}{rl}(x-3y{)}^2& =4xy\\ x^2-6xy+9y^2& =4xy\\ x^2-10xy+9y^2& =0\\ (x-y)(x-9y)& =0\end{array}$$ So ${\displaystyle \frac{x}{y}}=1$ or ${\displaystyle \frac{x}{y}}=9$. But from our restrictions we know that ${\displaystyle \frac{x}{y}}>3$, and so ${\displaystyle \frac{x}{y}}=9$.

2. Given that $m$ and $k$ are integers, find all values of $m$ and $k$ satisfying the equation $$9(7^k+7^{k+2})=5^{m+3}+5^m$$

   Solution

   We factor both sides of this equation to arrive at $$\begin{array}{rl}9(1+7^2)7^k& =5^m(1+5^3)\\ 9(50)7^k& =5^m(126)\\ 3^2\cdot 2\cdot 5^2\cdot 7^k& =5^m\cdot 2\cdot 3^2\cdot 7\end{array}$$ Now since both sides of this equation are products of primes, and integers have unique prime factorizations, it follows that $m=2$ and $k=1$ is the only solution.

3. Determine the points of intersection of the curves $y={\log }_{10}(x-2)$ and $y=1-{\log }_{10}(x+1)$.

   Solution

   Again the arguments of the logarithmic functions, $x-2$ and $x+1$, must be positive, which implies that $x>2$. Now $$\begin{array}{rl}{\log }_{10}(x-2)& =1-{\log }_{10}(x+1)\\ {\log }_{10}(x-2)+{\log }_{10}(x+1)& =1\\ {\log }_{10}[(x-2)(x+1)]& =1\\ (x-2)(x+1)& =10\\ x^2-x-2& =10\\ x^2-x-12& =0\\ (x-4)(x+3)& =0\end{array}$$ So $x=4$ or $x=-3$, but from our restriction $x>2$ and so $x=4$. The point of intersection is $(4,{\log }_{10}2)$ or $(4,1-{\log }_{10}5)$. Since ${\log }_{10}2+{\log }_{10}5=1$, these are equivalent answers.

4. Determine all values of $x$ such that ${\log }_2(9-2^x)=3-x$.

   Solution

   Once again the argument of the logarithm must be positive, implying that $9>2^x$. $$\begin{array}{rl}{\log }_2(9-2^x)& =3-x\\ 9-2^x=2^{3-x}& ={\displaystyle \frac{8}{2^x}}\end{array}$$

   Substituting $y=2^x$ we have $$\begin{array}{rl}9-y& ={\displaystyle \frac{8}{y}}\\ y^2-9y+8& =0\\ (y-1)(y-8)& =0\end{array}$$ Thus, $y=1$ or $y=8$. Since $y=2^x$, we obtain the corresponding values $x=0$ or $x=3$. Both of these values satisfy the restriction $9>2^x$ and so both are valid solutions.

5. The graph of $y=m^x$ passes through the points $(2,5)$ and $(5,n)$. What is the value of $mn$?

   Solution

   From the given information we have that $m^2=5$ and $n=m^5$. Thus, $m=\pm \sqrt{5}$ with corresponding values $n=(\pm \sqrt{5}{)}^5$. Therefore, $mn=(\sqrt{5}{)}^6=125$.

Problem Set

1. Determine the values of $x$ such that ${\log }_{x}2+{\log }_{x}4+{\log }_{x}8=1$.

2. Determine the values of $x$ such that ${12}^{2x+1}=2^{3x+7}\cdot 3^{3x-4}$.

3. What is the sum of the following series?

   $${\log }_{10}{\displaystyle \frac{3}{2}}+{\log }_{10}{\displaystyle \frac{4}{3}}+{\log }_{10}{\displaystyle \frac{5}{4}}+\cdots +{\log }_{10}{\displaystyle \frac{200}{199}}$$

4. Given that $x^3{y}^5=2^{11}\cdot 3^{13}$ and ${\displaystyle \frac{x}{y^2}}={\displaystyle \frac{1}{27}}$, solve for $x$ and $y$.

5. Given that ${\log }_{8}3=k$, express ${\log }_{8}18$ in the form $ak+b$ where $a$ and $b$ are rational numbers.

6. Determine the point(s) of intersection of the graphs of $y={\log }_2(2x)$ and $y={\log }_{4}x$.

7. The points $A(x_1,y_1)$ and $B(x_2,y_2)$ lie on the graph of $y={\log }_{a}x$. A horizontal line is drawn through the midpoint of $AB$ such that it intersects the curve at the point $C(x_3,y_3)$. Show that $(x_3{)}^2=x_1{x}_2$.

8. The graph of the function $y=a{x}^r$ passes through the points $(2,1)$ and $(32,4)$. Calculate the value of $r$.

9. Given that $2^{x+3}+2^x=3^{y+2}-3^y$ and $x$ and $y$ are integers, determine the values of $x$ and $y$.

10. Given that $f(x)=2^{4x-2}$, calculate, in simplest form, $f(x)\cdot f(1-x)$.

11. Find all values of $x$ such that ${\log }_5(x-2)+{\log }_5(x-6)=2$.

12. Let $x$ be a positive real number with $x\ne 1$. Prove that $a$, $b$, and $c$ are three numbers that form a geometric sequence if and only if ${\log }_{x}a$, ${\log }_{x}b$ and ${\log }_{x}c$ form an arithmetic sequence.

13. Determine all real values of $x$ for which $$3^{x+2}+2^{x+2}+2^x=2^{x+5}+3^x$$

14. Determine all real numbers $x$ for which $$({\log }_{10}x{)}^{{\log }_{10}({\log }_{10}x)}=10000$$

15. Determine all real numbers $x>0$ for which $${\log }_{4}x-{\log }_{x}16={\displaystyle \frac{7}{6}}-{\log }_{x}8$$