Indices, Surds and Logarithms, Worked Examples (easy)

Treat the number and each letter separately, and use the division law $a^{m}\div a^{n}=a^{m-n}$: when you divide powers of the same base, you subtract the indices. First divide the numbers: $12\div 3 = 4$.

Now the $x$ terms: $x^{2}\div x^{5}=x^{2-5}=x^{-3}$. Then the $y$ terms: $y^{5}\div y^{2}=y^{5-2}=y^{3}$.

Putting these together gives:

The question asks for positive indices only, so rewrite $x^{-3}$ as $\dfrac{1}{x^{3}}$, using $a^{-n}=\dfrac{1}{a^{n}}$. A negative index simply means the term belongs in the denominator:

So the simplified form is $\dfrac{4y^{3}}{x^{3}}$. A quick check with $x=1,\ y=1$: the original is $\dfrac{12}{3}=4$ and our answer is $\dfrac{4}{1}=4$, which agree, so the coefficient is right.

When both sides can be written as powers of the same base, the plan is to make the bases match and then equate the indices. The left side is already a power of $2$, so write the right side as a power of $2$ as well.

Since $32 = 2\times2\times2\times2\times2$, we have $32 = 2^{5}$.

The bases are now equal, so the indices must be equal. This is the key idea: if $a^{m}=a^{n}$ then $m=n$.

Setting the indices equal gives a simple linear equation:

So $x=2$. Check by substituting back: $2^{2+3}=2^{5}=32$, which matches the right-hand side, so the solution is correct.

You can only add surds once they share the same number under the root, so first simplify each one. The method is to pull out the largest perfect-square factor using $\sqrt{ab}=\sqrt{a}\,\sqrt{b}$.

For $\sqrt{12}$, the largest perfect square dividing $12$ is $4$, so $\sqrt{12}=\sqrt{4\times3}=\sqrt{4}\,\sqrt{3}=2\sqrt{3}$. For $\sqrt{27}$, the largest perfect square dividing $27$ is $9$, so $\sqrt{27}=\sqrt{9\times3}=\sqrt{9}\,\sqrt{3}=3\sqrt{3}$.

Both terms now carry the same surd $\sqrt{3}$, so treat $\sqrt{3}$ like a common factor and add the numbers in front: $2+3=5$.

So $\sqrt{12}+\sqrt{27}=5\sqrt{3}$. As a rough numerical check, $\sqrt{12}\approx3.46$ and $\sqrt{27}\approx5.20$, giving about $8.66$; and $5\sqrt{3}\approx5\times1.732=8.66$, so the answer is consistent.

A logarithm asks 'what power of the base gives this number?'. Read each term with the definition $\log_{a}b = c \Leftrightarrow a^{c}=b$, and work the two logs out separately.

For $\log_{2}16$: we need the power of $2$ that gives $16$. Since $2^{4}=16$, we have $\log_{2}16=4$.

For $\log_{3}27$: we need the power of $3$ that gives $27$. Since $3^{3}=27$, we have $\log_{3}27=3$.

So the value is $7$. Each step just turns the logarithm back into the index statement it stands for, which is the safest way to evaluate simple logs by hand.

Use the zero index law and the negative index law separately, then combine the results. The zero index law states that any non-zero base raised to the power $0$ equals $1$: $a^{0}=1$.

The negative index law states $a^{-n}=\dfrac{1}{a^{n}}$, so a negative index sends the term to the denominator: $2^{-3}=\dfrac{1}{2^{3}}=\dfrac{1}{8}$.

A surd should not be left in the denominator. Rationalise by multiplying both the numerator and the denominator by $\sqrt{3}$, which does not change the value of the fraction since $\dfrac{\sqrt3}{\sqrt3}=1$.

The denominator is now the whole number $3$, since $\sqrt{3}\times\sqrt{3}=3$. Divide the numerator by $3$ to finish simplifying.

A fractional index $a^{\frac{m}{n}}$ means take the $n$-th root of $a$, then raise the result to the power $m$: $a^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^{m}$. Here $n=3$ and $m=2$, so first find the cube root of $27$.

Now raise that cube root to the power $2$.

Neither $\log_4 96$ nor $\log_4 6$ is a whole number on its own, so evaluating them separately will not work here. Instead use the quotient law of logarithms, $\log_{a}x-\log_{a}y=\log_{a}\!\left(\dfrac{x}{y}\right)$, to combine them into a single logarithm first.

Now evaluate $\log_4 16$ directly from the definition: it asks what power of $4$ gives $16$. Since $4^{2}=16$, the value is $2$.