Quick questions on Indices and index laws for HSC Maths Advanced: the five index laws, zero and negative indices, fractional indices as roots and powers, simplifying index expressions, solving simple index equations by equating bases, and scientific notation

Because those are the only values that keep the division law true. Consider $\dfrac{a^3}{a^3}$. By ordinary cancelling it equals $1$.

Five laws let you combine powers of the same base. The first three combine two powers; the last two push a power across a product or quotient:

A fractional index brings in roots, and again the definition is chosen to keep the power-of-a-power law alive. Take $a^{1/2}$. The law says $\left(a^{1/2}\right)^2 = a^{(1/2) \times 2} = a^1 = a$, so $a^{1/2}$ is the number that squares to $a$: it is $\sqrt{a}$. The same reasoning gives $a^{1/3} = \sqrt[3]{a}$ and in general $a^{1/n} = \sqrt[n]{a}$, the $n$-th root.

When an index is both negative and fractional, deal with the pieces in a fixed order and each step stays simple. This single routine handles every numerical index question in the course, and it is the kind of explicit method the textbook leaves you to assemble yourself.

An equation such as $2^x = 32$ asks "what power of $2$ gives $32$?". The reliable method at this level is to write both sides as powers of the same base and then equate the indices, because a power is completely determined by its index once the base is fixed. Since $32 = 2^5$, the equation $2^x = 2^5$ forces $x = 5$. The same trick handles equations where the two sides use different-looking bases that are secretly powers of one common base: for $8^x = 4$, write $8 = 2^3$ and $4 = 2^2$, so $\left(2^3\right)^x = 2^2$ gives $2^{3x} = 2^2$ and hence $3x = 2$, that is $x = \dfrac{2}{3}$.

Scientific notation writes a number as $a \times 10^k$, where $1 \le a < 10$ (one non-zero digit before the decimal point) and $k$ is an integer. It is the natural home for the index laws, because multiplying and dividing such numbers just means handling the number parts and the powers of ten separately. To multiply, multiply the $a$ values and add the indices of $10$; to divide, divide the $a$ values and subtract the indices. After combining, you may need to renormalise so the front number again lies between $1$ and $10$: for instance $12 \times 10^8 = 1.2 \times 10^1 \times 10^8 = 1.2 \times 10^9$.