Logarithm Laws: See the Structure Before Calculating

Why multiplication becomes addition

Take two positive numbers that are powers of the same valid base: $M=b^u$ and $N=b^v$, where $b>0$ and $b\ne1$. Their product is $MN=b^{u+v}$. A logarithm asks for that exponent, so

This is the product rule. It works in both directions. You can expand one logarithm of a product into a sum, or condense a sum of same-base logarithms into one logarithm.

Notice what the rule does not say. $\log_b(M+N)$ does not generally equal $\log_b M+\log_b N$. For instance, with base 2 and $M=N=1$, the first expression is $\log_2 2=1$, while the proposed sum is $0+0=0$.

Before using factor tiles, predict the combined exponent. Keep every logged factor positive and use one common base. You cannot directly combine $\log_2 M+\log_3 N$ with this rule: the two terms measure exponents using different bases.

A denominator contributes a minus sign

Division of matching-base powers subtracts their exponents: $b^u/b^v=b^{u-v}$. The corresponding logarithm rule is

The base must again be positive and different from 1. Read the subtraction in order: numerator first, denominator second. Reversing the fraction changes the sign of its logarithm.

With several factors, organise the fraction before expanding. For positive $x,y,z,p,q$,

Both denominator factors receive minus signs. Writing the denominator as one parenthesised product helps: subtracting $\log_b(pq)$ subtracts the whole sum $\log_b p+\log_b q$.

Those positive-factor conditions matter. The number $(-6)/(-3)=2$ is positive, so its logarithm exists, but expanding it as $\log_b(-6)-\log_b(-3)$ creates undefined real logarithms. A positive combined argument alone does not justify every possible expansion.

A power comes down—but keep the domain

For a positive number $x$, a valid base $b$ and any real exponent $r$,

To see why, let $u=\log_b x$, so $x=b^u$. Then $x^r=b^{ur}$, whose logarithm is $ur$. The rule handles positive, negative and fractional powers: a square root contributes a factor $1/2$, while a reciprocal contributes a factor $-1$.

The condition $x>0$ is essential for that general statement. A square creates a useful special case. For any real $x\ne0$, the argument $x^2$ is positive, even when $x$ is negative. The form that preserves its whole domain is

Here $|x|$ is the nonnegative magnitude, or absolute value (निरपेक्ष मान). Because $x\ne0$, it is strictly positive. Writing $2\log_b x$ instead would wrongly discard every negative $x$. A simpler-looking expression is only equivalent where its logarithms still exist.

A power on the base divides the answer

A power on the base behaves differently from a power on the argument. With $a>0$, $a\ne1$, $x>0$ and real $n\ne0$,

Let $t=\log_{a^n}x$. By definition, $(a^n)^t=x$, or $a^{nt}=x$. Thus $nt=\log_a x$, and dividing by $n$ gives the rule. If $n=0$, the new base would be $a^0=1$, which is not allowed.

When both positions carry powers, combine the two operations:

The conditions are unchanged, with $m$ any real number. A negative $n$ is allowed; it makes the new base a reciprocal power. A zero $m$ is allowed too: the argument becomes 1 and the logarithm becomes zero.

Use the two power controls separately before changing both. Say what changed out loud: “the argument power multiplies; the base power divides”. This helps prevent applying the argument rule to the small subscript.

Let prime factors carry the arithmetic

When several same-base logarithms contain fractions, a dependable route is condense, factor, cancel, expand. Condense only after confirming that every original argument is positive. Then use exact fractions rather than rounded decimals.

Prime factorisation turns large-looking numbers into a short exponent record. For example, $125=5^3$, $81=3^4$, $64=2^6$ and $243=3^5$. Equal prime factors cancel between numerator and denominator by subtracting exponents.

The remaining exponents become logarithm coefficients. A prime in the denominator produces a negative coefficient. A prime that cancels completely has coefficient zero; you should not invent a leftover term just because the question mentions it.

Keep the common base visible throughout. The arithmetic below works for any $b>0$, $b\ne1$, not only base 10. Once you have an answer, check it backwards: turn the coefficients into powers and reconstruct the fraction. That inverse check catches both dropped factors and reversed signs.