Expanding and condensing logs

A log is an exponent, so log properties are exponent rules read backwards. Multiplying powers adds exponents, so the log of a product adds logs. Dividing powers subtracts exponents. A power of a power multiplies exponents, so an exponent inside a log slides out front as a factor:

1. Product: $\log_b(xy) = \log_b x + \log_b y$
2. Quotient: $\log_b\!\left(\tfrac{x}{y}\right) = \log_b x - \log_b y$
3. Power: $\log_b(x^k) = k\,\log_b x$

A quick numeric check keeps them honest: $\log_2(8 \cdot 4) = \log_2 32 = 5$, and $\log_2 8 + \log_2 4 = 3 + 2 = 5$. Any claimed property you cannot verify with small powers of 2 is not a property.

Notice what the three rules act on: a product, a quotient, or a power inside one log. Nothing acts on a sum. There is no rule for $\log_b(x + y)$; it does not split, shrink, or simplify. Test it: $\log_2(8 + 8) = \log_2 16 = 4$, but $\log_2 8 + \log_2 8 = 6$. Not equal, not close, not fixable.

Equally fake: turning $\log_b x + \log_b y$ into $(\log_b x)(\log_b y)$, or $\log_b x - \log_b y$ into $\frac{\log_b x}{\log_b y}$. The product property says a SUM of logs condenses to the log of a product; it never says logs multiply. Keep the direction straight: the operation moves DOWN a level (multiplication inside becomes addition outside), exactly as exponents demote multiplication to addition.

Expanding breaks one log into pieces, useful for isolating a variable: $\log\!\left(\tfrac{x^2 y}{z}\right) = 2\log x + \log y - \log z$. Work outside-in: quotient first, then product, then powers.

Condensing runs the same rules in reverse, packing a sum into a single log so an equation can be exponentiated: $2\log x + \log y = \log(x^2 y)$. The discipline that saves points: coefficients must become exponents BEFORE terms merge. $2\log x + \log y$ is $\log(x^2y)$, not $\log(2x + y)$ and not $2\log(xy)$.

One quiet caveat: expansions are equalities on the domain where every piece is defined. $\log(x^2)$ exists for negative x but $2\log x$ does not, so the rewrite narrows the domain to x > 0. The AP exam mostly keeps you on safe ground here, but noting the restriction is what full reasoning looks like.

Calculators own base 10 and base e, so every other base gets there by change of base:

$$\log_b x = \frac{\log x}{\log b}$$

So $\log_5 40 = \log 40 / \log 5 \approx 1.602/0.699 \approx 2.29$, and the check $5^{2.29} \approx 40$ confirms it.

Keep this formula visually separate from the quotient property, because they are mirror-image statements. Change of base is a quotient of two logs, $\frac{\log 40}{\log 5}$, and it does NOT equal $\log(40/5)$. The quotient property is a log of a quotient, $\log(40/5) = \log 40 - \log 5$, and it does NOT equal $\frac{\log 40}{\log 5}$. Students who blur the two lose the point in both directions.