Properties of Logarithms: Practice

Change every factor to base $2$. Each one becomes a fraction whose numerator cancels the denominator of the next.

$\log_3 4 = \frac{\log_2 4}{\log_2 3}, \qquad \log_4 5 = \frac{\log_2 5}{\log_2 4}, \qquad \log_5 8 = \frac{\log_2 8}{\log_2 5}$

Multiplying the chain, every intermediate logarithm cancels and only the ends survive.

$\log_2 3 \cdot \frac{\log_2 4}{\log_2 3} \cdot \frac{\log_2 5}{\log_2 4} \cdot \frac{\log_2 8}{\log_2 5} = \log_2 8 = 3$

Expand with the quotient and power rules, treating the square root as the exponent $\tfrac{1}{2}$.

$\log_b\!\left(\frac{x^2}{\sqrt{y}}\right) = 2\log_b x - \frac{1}{2}\log_b y$

Substitute the two given values, and watch the sign: subtracting a negative adds.

$= 2(3) - \frac{1}{2}(-2) = 6 + 1 = 7$

A negative logarithm is nothing exotic. It just means $y = b^{-2}$, which is a positive number less than $1$, so $\sqrt{y}$ is perfectly legal.

Change of base with $a = 6$ turns each reciprocal into a single logarithm: $\dfrac{1}{\log_2 6} = \log_6 2$ and $\dfrac{1}{\log_3 6} = \log_6 3$.

Now the product rule collapses the sum.

$\log_6 2 + \log_6 3 = \log_6(2 \cdot 3) = \log_6 6 = 1$

The reciprocal identity $\log_b M = \dfrac{1}{\log_M b}$ is what makes the flip legal, and it needs $M \ne 1$, which holds here.

Split the quotient first, then the product on top, writing $\sqrt{y} = y^{1/2}$.

$\log_b\!\left(\frac{x^3 \sqrt{y}}{z^2}\right) = \log_b\!\left(x^3\right) + \log_b\!\left(y^{1/2}\right) - \log_b\!\left(z^2\right)$

The power rule now brings each exponent to the front.

$= 3\log_b x + \frac{1}{2}\log_b y - 2\log_b z$

Only the denominator is subtracted, and the square root becomes the coefficient $\tfrac{1}{2}$, not $2$.

The right side contains $\log_5(-x)$, which exists only when $-x > 0$, that is when $x < 0$. So no positive $x$ can work, and $x = 0$ fails on both sides.

For $x < 0$ the number $-x$ is positive, and $(-x)^2 = x^2$, so the power rule applies to $-x$ and gives equality.

$2\log_5(-x) = \log_5\!\left((-x)^2\right) = \log_5\!\left(x^2\right)$

Check at $x = -5$: the left side is $\log_5 25 = 2$, and the right side is $2\log_5 5 = 2$. This is the mirror image of the familiar rule $\log_5\!\left(x^2\right) = 2\log_5 x$, which holds exactly when $x > 0$. Both are special cases of $\log_5\!\left(x^2\right) = 2\log_5 |x|$.

The trick is to write $5$ in terms of numbers whose logarithms you already know: $5 = \dfrac{10}{2}$.

The quotient rule then does the rest, using $\log_{10} 10 = 1$.

$\log_{10} 5 = \log_{10} 10 - \log_{10} 2 \approx 1 - 0.3010 = 0.6990$

This single relationship is why an old four-figure table needed only the primes: everything else was built from them.

Change base from $\sqrt{b}$ to $b$, which is the base you have information about.

$\log_{\sqrt{b}} 3 = \frac{\log_b 3}{\log_b \sqrt{b}}$

The denominator is easy: $\sqrt{b} = b^{1/2}$, so $\log_b \sqrt{b} = \tfrac{1}{2}$.

$= \frac{0.6}{\tfrac{1}{2}} = 1.2$

Shrinking the base makes the logarithm larger, which fits: it takes a bigger exponent on $\sqrt{b}$ than on $b$ to reach the same number.

Fold the coefficient back inside the logarithm with the power rule before doing anything else.

$2\log_b 3 = \log_b\!\left(3^2\right) = \log_b 9$

Now the exponential and the logarithm are inverses, so they cancel.

$b^{\,\log_b 9} = 9$

The answer does not depend on $b$ at all, which is the signature of an inverse pair undoing each other.

Change the base of $\log_x y$ to $b$, since $b$ is the base you have values in.

$\log_x y = \frac{\log_b y}{\log_b x} = \frac{5}{2}$

A sanity check with real numbers: take $b = 10$, so $x = 10^2 = 100$ and $y = 10^5$. Then $x^{5/2} = \left(10^2\right)^{5/2} = 10^5 = y$, confirming $\log_x y = \tfrac{5}{2}$.

Write $3.5$ using only the numbers you have: $3.5 = \dfrac{7}{2}$.

The quotient rule turns that division into a subtraction.

$\log_{10} 3.5 = \log_{10} 7 - \log_{10} 2 \approx 0.845 - 0.301 = 0.544$

Check the size: $3.5$ lies between $1$ and $10$, so its base-10 logarithm must lie between $0$ and $1$, which rules out $1.146$ immediately.

The square root covers the entire fraction, so the power rule brings a $\tfrac{1}{2}$ out in front of everything.

$\log_b \sqrt{\frac{x}{y^3}} = \frac{1}{2}\log_b\!\left(\frac{x}{y^3}\right) = \frac{1}{2}\left(\log_b x - 3\log_b y\right)$

Now distribute the $\tfrac{1}{2}$ across both terms, which is the step most often dropped.

$= \frac{1}{2}\log_b x - \frac{3}{2}\log_b y$

Split the quotient, then the product, writing $\sqrt{y} = y^{1/2}$.

$\log_{10}\!\left(\frac{100x^3}{\sqrt{y}}\right) = \log_{10} 100 + \log_{10}\!\left(x^3\right) - \log_{10}\!\left(y^{1/2}\right)$

Now $\log_{10} 100 = 2$ because $10^2 = 100$, and the power rule handles the other two terms.

$= 2 + 3a - \frac{1}{2}c$

The constant $100$ contributes its logarithm, $2$, not the number $100$ itself.