Introduction to Logarithms: Free Response

Ask the defining question for the first one: six to the what gives two hundred sixteen? Running up the powers of $6$ answers it directly.

$6^3 = 216 \quad\Longrightarrow\quad \log_6 216 = 3.$

The second is not a whole-number power, so rewrite both $32$ and $8$ as powers of the common smaller number $2$.

$32^{y} = 8 \quad\Longrightarrow\quad 2^{5y} = 2^{3} \quad\Longrightarrow\quad 5y = 3 \quad\Longrightarrow\quad y = \frac{3}{5}.$

Test $b = 1$ first. Since $1^{y} = 1$ no matter what $y$ is, the equation $1^{y} = 5$ has no solution and $1^{y} = 1$ has infinitely many, so $1^{x}$ is not one-to-one and has no inverse to name.

$1^{y} = 1 \text{ for every } y \quad\Longrightarrow\quad \log_1 x \text{ cannot be defined.}$

Test $b = -9$ next. A power such as $(-9)^{1/2}$ is not a real number at all, so $(-9)^{x}$ is not even a function on the real numbers, and there is nothing there to invert.

Test $b = 0.4$. It is positive and not equal to $1$, so $0.4^{x}$ is a genuine one-to-one exponential with a genuine inverse.

$b = 0.4 \text{ is the only allowed base of the three.}$

Both facts come from one substitution into the defining equation $b^{y} = x$.

Substitute $x = 1$: since $b^{0} = 1$ for every allowed base,

$b^{0} = 1 \quad\Longrightarrow\quad \log_b 1 = 0.$

Substitute $x = b$: since $b^{1} = b$ by definition of a first power,

$b^{1} = b \quad\Longrightarrow\quad \log_b b = 1.$

The two look alike on the page because each pairs a special input, either $1$ or the base itself, with a small whole-number output, either $0$ or $1$; nothing about the symbols reminds you which special input goes with which output.

Since each round halves the field, the number of rounds needed for $512$ players is the exponent that turns $2$ into $512$.

$2^{9} = 512 \quad\Longrightarrow\quad \log_2 512 = 9.$

Nine rounds exactly seat $512$ players with no byes needed anywhere.

List the powers of $2$ nearest to $200$: $2^{7} = 128$ and $2^{8} = 256$.

$128 < 200 < 256.$

Applying $\log_2$ preserves the ordering because $\log_2$ is increasing, and $\log_2 128 = 7$, $\log_2 256 = 8$:

$7 < \log_2 200 < 8.$

Seven rounds can only seat $2^{7} = 128$ players, fewer than the $200$ entrants, so seven rounds is not enough. The bracket must run a full eighth round, with byes handed out in the first round to make the numbers work.

A bracket run for $k$ rounds can seat at most $2^{k}$ players, since each round only doubles the capacity of the round before it.

$\text{capacity after } k \text{ rounds} = 2^{k}.$

When the field size is not an exact power of two, its logarithm is not a whole number, and rounding to the nearer integer answers a different question, how close the field size is to a clean power of two, not whether every entrant has a seat. If the lower integer is used because the logarithm sits nearer to it, the resulting $2^{k}$ can still be smaller than the actual field, leaving entrants with nowhere to play. Only rounding up, the ceiling of the logarithm, guarantees $2^{k}$ is at least the number of entrants, whatever the field size happens to be.

When the field size IS an exact power of two, as in part A, its logarithm already lands exactly on a whole number, so there is no gap to round across: that integer already is the exact number of rounds needed, neither more nor less.

The first expression matches the pattern $\log_b\!\left(b^{t}\right)$, which cancels to the exponent itself for every real $t$.

$\log_5\!\left(5^{-9}\right) = -9.$

The second matches the pattern $b^{\log_b x}$, which cancels to $x$ whenever $x > 0$.

$6^{\log_6 41} = 41.$

Rewrite the outer base as a power of $2$, matching the base of the logarithm inside the exponent.

$16^{\log_2 3} = \left(2^{4}\right)^{\log_2 3} = \left(2^{\log_2 3}\right)^{4}.$

The cancellation law $b^{\log_b x} = x$ now collapses the inner power, since its base matches the logarithm's own base.

$\left(2^{\log_2 3}\right)^{4} = 3^{4} = 81.$

Look at what each expression asks the logarithm to accept.

In $b^{\log_b x} = x$, the inner piece $\log_b x$ has to exist before the outer power even makes sense, and $\log_b x$ is only defined when its input $x$ is positive.

$x \le 0 \quad\Longrightarrow\quad \log_b x \text{ does not exist} \quad\Longrightarrow\quad b^{\log_b x} \text{ is meaningless.}$

In $\log_b\!\left(b^{t}\right) = t$, the inner piece $b^{t}$ is a power of a positive base, so it is positive for every real $t$ whatsoever, positive, negative, or zero. Its input is never in question, so the equation needs no fine print at all.

Switch to exponential form: the base raised to the second power gives $49$.

$x^{2} = 49.$

The real roots are $x = 7$ and $x = -7$, but a logarithm base must be positive and different from $1$, so only $x = 7$ survives as an actual base.

$x = 7.$

The unknown sits in the exponent, so switch straight to logarithmic form.

$x = \log_9 \frac{1}{729}.$

Since $729 = 9^{3}$, a reciprocal calls for a negative exponent.

$9^{-3} = \frac{1}{9^{3}} = \frac{1}{729} \quad\Longrightarrow\quad x = -3.$

Switch to exponential form: the argument equals $2^{1}$.

$x^{2} + x = 2 \quad\Longrightarrow\quad x^{2} + x - 2 = 0 \quad\Longrightarrow\quad (x - 1)(x + 2) = 0,$

which gives $x = 1$ or $x = -2$.

Check the argument for each root: at $x = 1$ it is $1 + 1 = 2$, and at $x = -2$ it is $4 - 2 = 2$. Both are positive, so both are genuine solutions.

This was never in doubt, because of how the equation was built. For an equation of this same shape, a logarithm of a polynomial expression in $x$ set equal to a fixed constant $k$, switching $\log_b(\text{polynomial}) = k$ into exponential form always sets the polynomial equal to $b^{k}$, a positive number. Every algebraic root of the resulting polynomial equation therefore makes the argument equal to that same positive value, so the domain check can never throw out a root of an equation built this same way, whatever the coefficients happen to be.

Reflecting a point across the line $y = x$ swaps its two coordinates, so apply that swap to all three given points at once.

$\left(-1,\ \tfrac19\right) \to \left(\tfrac19,\ -1\right), \qquad (0,\ 1) \to (1,\ 0), \qquad (2,\ 81) \to (81,\ 2).$

Each feature of the logarithm is the matching feature of the exponential, with the roles of input and output swapped.

The exponential has domain all real numbers and range $y > 0$; swapped, the logarithm has

$\text{domain } x > 0, \qquad \text{range all real numbers}.$

The exponential's horizontal asymptote $y = 0$ becomes, after the swap, a vertical line for the logarithm.

$x = 0.$

Take any two points on $y = 7^{x}$ with the first lower and further left than the second, since the curve is increasing: for instance $A = (0,\ 1)$ and $B = (1,\ 7)$, where $A$ sits lower and to the left of $B$. Reflecting across $y = x$ swaps each point's coordinates, turning $A$ into $A' = (1,\ 0)$ and $B$ into $B' = (7,\ 1)$. Compare the two swapped points directly: $A'$ still has the smaller $x$-coordinate, since $1 < 7$, so $A'$ is still to the left of $B'$, and $A'$ still has the smaller $y$-coordinate, since $0 < 1$, so $A'$ is still lower than $B'$. The swap does not reverse which point is lower and further left; it preserves that order.

$\text{lower, further left} \ \longrightarrow \ \text{after reflection} \ \longrightarrow \ \text{still lower, further left}.$

That order-preservation is exactly the shape of an increasing curve read left to right, so the reflected graph, which is $y = \log_7 x$, is increasing too.

For the second fact, every exponential $y = b^{x}$ passes through $(0,\ 1)$, since $b^{0} = 1$ regardless of the base. Reflecting that single point across $y = x$ swaps its coordinates to $(1,\ 0)$, and the swap does not depend on which base $b$ was chosen, so it happens for every logarithm's graph alike.