Logarithms

Two forms, one fact. The exponential form and the logarithmic form describe the same relationship between a base, an index, and a result:

where: $a$ is the base, $x$ is the index (the power), and $N$ is the result. Read $\log_a N=x$ as "the power to which $a$ must be raised to give $N$ is $x$." The condition $a>0,\ a\neq1$ keeps the base sensible (a base of $1$ or a negative base breaks the pattern), and $N>0$ because a positive base raised to any real power stays positive.

Three laws, one source. Since a logarithm is an index, the three laws of indices translate directly into three laws of logarithms:

where: $M$ and $N$ are positive numbers, and $k$ is any real index. The product law turns multiplication inside a log into addition outside it; the quotient law turns division into subtraction; the power law brings an exponent down as a multiplying coefficient.

Two anchor values. Two further results follow straight from the definition and are worth memorising cold:

$\log_a a=1$ because $a^1=a$; $\log_a 1=0$ because $a^0=1$ for any valid base $a$.

The three logarithm laws are not separate facts to memorise; each one is a direct translation of the corresponding law of indices, because $\log_a N$ is, by definition, the index to which $a$ must be raised to produce $N$.

Step 1. Let $M=a^x$ and $N=a^y$ for some base $a$, so that by definition $\log_a M=x$ and $\log_a N=y$ — this is just restating what the two logarithms mean as indices.

Step 2. Multiply $M$ and $N$ using the multiplication law of indices, $a^x\times a^y=a^{x+y}$, so that $MN=a^{x+y}$; converting this back to logarithmic form gives $\log_a(MN)=x+y$, and substituting $x=\log_a M$, $y=\log_a N$ gives the product law $\log_a(MN)=\log_a M+\log_a N$.

Step 3. Divide instead of multiply: the division law of indices gives $\dfrac{M}{N}=\dfrac{a^x}{a^y}=a^{x-y}$, so $\log_a\!\left(\dfrac{M}{N}\right)=x-y=\log_a M-\log_a N$, which is the quotient law.

Step 4. Raise $M=a^x$ to a power $k$: the power law of indices gives $M^k=(a^x)^k=a^{xk}$, so $\log_a(M^k)=xk=k\log_a M$, which is the power law. All three logarithm laws therefore come from the same three laws of indices you already know — nothing new is being assumed, only restated in logarithmic language.

If $y=\dfrac{p^3\sqrt{q}}{r^2}$, express $\log y$ in terms of $\log p$, $\log q$, and $\log r$; then use $\log_5 125=3$ to check the definition still holds for a numerical case.

Step 1. Rewrite the square root as a fractional index, so the power law can be applied cleanly later:

Step 2. Apply the quotient law to split the fraction into a difference of two logarithms:

Step 3. Apply the product law to the first term, separating the product $p^3q^{1/2}$:

Step 4. Apply the power law to bring each index down as a coefficient:

So $\boxed{\log y = 3\log p+\dfrac{1}{2}\log q-2\log r}$.

Competitive stretch

The same definition handles pure numbers directly, without any expansion. Since $5^3=125$, the equivalence $a^x=N\Leftrightarrow\log_a N=x$ gives $\log_5 125=3$ immediately — check: $5\times5\times5=125$. And reversing it, $\log_2 32=5$ converts back to $2^5=32$, since $2\times2\times2\times2\times2=32$. Working through worked example 2 one step at a time — quotient, then product, then power — rather than trying to expand everything in one jump, is what keeps three laws from tangling together when an expression uses all of them at once.

It helps to stop thinking of "logarithms" as a brand-new mathematical object and instead see them as a change of viewpoint on something you already know: indices. Every statement you can write with an exponent has a matching statement you can write with a logarithm, and vice versa. When a question gives you $y=\dfrac{a^4b^2}{c^3}$ and asks for $\log y$, it is really asking you to track what happens to each factor's index as the product, quotient, and power laws peel the expression apart — the logarithm notation is just bookkeeping for those indices once they are written as coefficients. This viewpoint also explains why the syllabus stops where it does. Class 9 asks you to expand and simplify, never to solve equations like $\log_2 x=5$ for a variable buried inside a more complex expression, and never to change between different bases. Both of those skills need extra machinery — solving needs you to treat the logarithm as an invertible function, and changing base needs a formula that is not part of this chapter. Staying disciplined about scope means you will not lose marks trying to prove the laws from scratch in an exam that only asks you to apply them.

• Confusing $\log_a(MN)$ with $\log_a M\times\log_a N$ — the product law converts multiplication inside the log into addition outside it, never multiplication.
• Writing $\log_a(M+N)$ as $\log_a M+\log_a N$ — there is no law for the log of a sum; the three laws apply only to products, quotients, and powers.
• Forgetting to convert a root like $\sqrt{q}$ into the fractional index $q^{1/2}$ before applying the power law, and leaving it un-expanded instead.
• Mixing up $\log_a a=1$ with $\log_a 1=0$ under exam pressure — anchor them to $a^1=a$ and $a^0=1$ respectively rather than memorising them as bare facts.
• Trying to apply the quotient law to $\log_a(M-N)$, treating subtraction inside a log the same as division — the quotient law needs a genuine fraction $M/N$, not a difference.

Ask the tutor: Want to practice expanding a messier expression like $\log\left(\dfrac{a^4b^2}{c^3}\right)$ or convert a batch of exponential statements into log form — ask your AI tutor to walk through a few with you, step by step.