Laws of Exponents with Rational Exponents
Rational Exponents and Roots
For any positive real number $a$ and rational number $\frac{m}{n}$ where $m$ and $n$ are integers and $n \neq 0$:
$$a^{\frac{m}{n}} = \sqrt[n]{a^m}$$
Let's evaluate $8^{\frac{2}{3}}$:
$$8^{\frac{2}{3}} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4$$
Here, we first square 8, then take the cube root of the result.
- When $n$ is even, $a^{\frac{1}{n}}$ refers specifically to the positive $n$th root of $a$.
- This convention ensures that exponential expressions with rational exponents are well-defined for positive real bases.
Properties of Exponents
- The laws of exponents are as follows:
- Product rule: $a^x \cdot a^y = a^{x+y}$
- Quotient rule: $\frac{a^x}{a^y} = a^{x-y}$
- Power rule: $(a^x)^y = a^{xy}$
- These rules apply regardless of whether $x$ and $y$ are integers, rational, or real numbers.
Evaluate $27^{\frac{4}{3}}$.
Solution
$$27^{\frac{4}{3}} = 27^{\frac{1}{3} \cdot 4} = \sqrt[3]{27}^4 = 3^4 = 81$$
We applied the power rule, then simplified the rational exponent, and finally evaluated the result.
- Students often confuse $a^{\frac{m}{n}}$ with $\frac{a^m}{a^n}$.
- Remember, $a^{\frac{m}{n}}$ is a single number raised to a fractional power, while $\frac{a^m}{a^n}$ is a quotient of two exponential expressions that evaluates to $a^{m - n}$.
Laws of Logarithms
Logarithms are the inverse operations of exponents, and they follow a set of laws that mirror the properties of exponents.
Basic Logarithm Laws
- The basic logarithm laws can be derived from the exponent laws.
- For positive real numbers $a$, $x$, and $y$, where $a \neq 1$:
- Product rule: $\log_a(xy) = \log_a(x) + \log_a(y)$
- Quotient rule: $\log_a(\frac{x}{y}) = \log_a(x) - \log_a(y)$
- Power rule: $\log_a(x^m) = m \log_a(x)$
Simplify $\log_2(16\sqrt2)$.
Solution
$$\log_2(16\sqrt2) = \log_2(16)+\log_2(\sqrt2) = 4+\frac12 = \frac92$$
We applied the product rule in reverse, then evaluated the logarithm.
- When simplifying logarithmic expressions, look for opportunities to apply these laws in either direction.
- Sometimes combining logarithms is helpful, while other times separating them is more useful.
Additional Logarithm Properties
- $\log_a(1) = 0$ for any base $a > 0$, $a \neq 1$
- $\log_a(a) = 1$ for any base $a > 0$, $a \neq 1$
- $a^{\log_a(x)} = x$ for $x > 0$ and $a > 0$, $a \neq 1$
These properties are often useful in solving logarithmic and exponential equations.
- The domain of $\log_a(x)$ is all positive real numbers ($x > 0$).
- The base $a$ must also be positive and not equal to 1.
Change of Base Formula for Logarithms
- The change of base formula allows us to express a logarithm with any base in terms of logarithms with a different base.
- This is particularly useful when you have expressions with logarithms in different bases, as it allows you to simplify the equation further.
- For positive real numbers $a$, $b$, and $x$, where $a \neq 1$ and $b \neq 1$: $$\log_a(x) = \frac{\log_b(x)}{\log_b(a)}$$
- To see why this works, let's consider $a$, $b$, $x$, and $y$ where $y = \log_a(x)$, and $b$ is an arbitrary positive number.
- We can also write this as $a^y = x$.
- Taking the logarithm in base $b$ of both sides, we get $\log_b(a^y) = \log_b(x)$.
- Simplifying, we get $y\log_b(a) = \log_b(x)$.
- Substituting $y = \log_a(x)$ and dividing both sides by $\log_b(a)$, we get: $$\log_a(x) = \frac{\log_b(x)}{\log_b(a)}$$
Simplify $\log_3(4) + \log_9(64)$.
Solution
$$=\log_3(2^2) + \frac{\log_3(2^6)}{\log_3(9)}$$
$$=2\log_3(2) + \frac{6\log_3(2)}{2}$$
$$=2\log_3(2) + 3\log_3(2)$$
$$=5\log_3(2)$$
When using a calculator, it's often easiest to use either natural logarithms ($\ln$) or common logarithms ($\log$) in the change of base formula.
Solving Exponential Equations
- Exponential equations are equations where the variable appears in the exponent, in the form $a^f(x)$.
- Solving these equations often involves using logarithms to "bring down" the exponent.
General Approach
- Isolate the exponential term on one side of the equation.
- Take the logarithm of both sides (any base will work, but look for what the base of the exponential term is).
- Use logarithm laws to simplify.
- Solve for the variable.
Solve the equation $2^{x-1} = 10$.
Solution
- The exponential term is already isolated.
- Take the natural log of both sides: $$\log_2(2^{x-1}) = \log_2(10)$$
- Simplify the left-hand side: $$x-1 = \log_2(10)$$
- Solve for $x$: $$x = \log_2(10) + 1 \approx 4.32$$
Some exponential equations may require additional algebraic manipulation before applying logarithms.
Solve the equation $(\frac{1}{3})^x = 9^{x+1}$.
Solution
- Rewrite the right side: $$(\frac{1}{3})^x = (3^2)^{x+1}$$
- Simplify the right side: $$(\frac{1}{3})^x = 3^{2x+2}$$
- Rewrite the left side: $$(3^{-1})^x = 3^{2x+2}$$
- Simplify the left side: $$3^{-x} = 3^{2x+2}$$
- Taking $\log_3$ of both sides: $$-x = 2x+2$$
- Solve for $x$: $$-3x = 2 \qquad x = -\frac{2}{3}$$
- When solving exponential equations, if both sides have the same base, then the exponents are equal (i.e. if $a^x = a^y$, $x = y$.)
- This can save you a bit of time on a test.
Solve the following exponential equations (in order of difficulty):
- $5^{2x+1} = 3^{x}$
- $7^{x^2} = 49^{6 - 2x}$
- $2^{5^{2x}} = 5^{2^{5x}}$
