Work with indices and exponential equations

Keep the index laws and their domains in view

Identify the bases before choosing a law. Multiplying powers with the same base adds exponents; raising a power to another power multiplies them.

For a positive integer n, $a^n$ means n factors of a multiplied together; a is the base and n the index or exponent. The source groups its nine laws as follows.

Same base:

$$a^ma^n=a^{m+n},\qquad\frac{a^m}{a^n}=a^{m-n},\qquad(a^m)^n=a^{mn}.$$

Same index:

$$a^mb^m=(ab)^m,\qquad\frac{a^n}{b^n}=\left(\frac ab\right)^n.$$

Zero, negative and fractional indices:

$$\begin{aligned} a^0&=1,\\ a^{-m}&=\frac1{a^m},\\ a^{1/m}&=\sqrt[m]a\quad(m\in\mathbb Z^+),\\ a^{m/n}&=\sqrt[n]{a^m}=(\sqrt[n]a)^m\quad(n\in\mathbb Z^+). \end{aligned}$$

The source lists these compactly without all domain qualifications. For unrestricted real exponents, use positive bases. Denominators and negative powers require nonzero bases; the zero-power rule above excludes zero. A real even root requires a nonnegative radicand.

For negative bases, rational exponents need a specified real-root convention and cannot be manipulated as unrestricted real exponents. Example 7(b) uses thirds: interpret them via the real cube root, with $x\ne0$. Continue with Tutorial Question 8.

Solve exponential equations with equal bases or a substitution

First try to express both sides using the same base. If several powers of one exponential appear, substitute for that exponential and retain its positive range.

For $a>0$ and $a\ne1$, the exponential is one-to-one: $a^x=a^n$ implies $x=n$. The source graphs show an increasing curve for a>1 and a decreasing curve for 0<a<1. Both pass through (0,1), stay positive and approach the x-axis without reaching zero. Their domain is all real numbers and range is positive real numbers.

Example 8(a) is the same printed equation as Tutorial Question 9. The lecture working writes $9=3^2$ and $27=3^3$:

$$\begin{aligned} (3^2)^{2x+1}-(3^3)^x&=0,\\ 3^{4x+2}-3^{3x}&=0,\\ 3^{4x+2}&=3^{3x},\\ 4x+2&=3x,\\ x&=-2. \end{aligned}$$

The tutorial solution also gives a factorisation argument. Example 8(b) introduces $y=2^x>0$ and rejects a negative quadratic candidate; the new exercise keeps all its printed stages.