Introduction to Logarithms: Free Response

Reaching $2$ from a base of $4$ takes a square root, since $4^{1/2}=\sqrt{4}=2$:

$4^{1/2} = 2 \;\Rightarrow\; \log_{4}(2) = \frac{1}{2}.$

A reciprocal always takes a negative exponent, and $2^{-5} = \dfrac{1}{2^{5}} = \dfrac{1}{32}$:

$2^{-5} = \frac{1}{32} \;\Rightarrow\; \log_{2}\!\left(\frac{1}{32}\right) = -5.$

The exponential $5^{x}$ is positive for every real exponent $x$, whatever sign $x$ itself carries: its range is

$y > 0.$

A logarithm only ever reports an exponent the exponential could actually have produced, so its input has to lie in that range. Neither $-4$ nor $0$ does, so

$\log_{5}(-4) \text{ and } \log_{5}(0) \text{ are undefined.}$

The number $0.02$ is positive, so it is a legitimate input, and $\log_{5}(0.02)$ is defined (it comes out negative, since $0.02$ is smaller than $1$).

Pick a base and a specific positive pair, and compute each side completely on its own rather than assuming they will agree.

With base $2$ and $M=N=8$, the left side adds the two inputs first, then takes the logarithm:

$\log_{2}(8+8) = \log_{2}(16) = 4,$

since $2^{4}=16$. The right side takes the two logarithms first, then adds:

$\log_{2}(8) + \log_{2}(8) = 3 + 3 = 6,$

since $2^{3}=8$. The two sides give $4$ and $6$, which disagree, so the claim fails on this pair.

A claim made for ALL positive $M$ and $N$ is destroyed by one pair that breaks it. But refuting the unrestricted claim does not hand over a replacement formula; it only rules out this particular shortcut. The product law, which combines $\log_b(MN)$ rather than $\log_b(M+N)$, is the rule that actually governs a product.

The common log carries no base at all and means base $10$: since $10^{3}=1000$,

$\log(1000) = 3.$

The natural log is written $\ln$ and means base $e$; it undoes $e$ raised to a power directly,

$\ln(e^{4}) = 4.$

Since $y=\log_{5}(x)$ is the inverse of $y=5^{x}$, each feature is the mirror of one of the exponential's own features, 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, the logarithm's vertical asymptote

$x = 0.$

The exponential passes through $(0,1)$, so swapped, the logarithm passes through $(1,0)$.

Both facts come from a single substitution into the definition $b^{x}=y$.

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

$b^{0} = 1 \;\Rightarrow\; \log_{b}(1) = 0.$

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

$b^{1} = b \;\Rightarrow\; \log_{b}(b) = 1.$

The two are easy to mix up precisely because they look alike on the page: each pairs a special input (either $1$ or the base $b$ itself) with a small whole-number output (either $0$ or $1$), and nothing about the symbols themselves reminds you which special input goes with which output.

Substitute $c=10^{-3}$ into the formula and evaluate the logarithm directly from the definition:

$\text{pH} = -\log(10^{-3}) = -(-3) = 3,$

since $10$ raised to the power $-3$ is $10^{-3}$, so $\log(10^{-3}) = -3$.

Rearrange the formula first, then convert straight to exponential form using the definition of a logarithm (no law is needed for this step):

$\text{pH} = -\log(c) = 5.6 \;\Rightarrow\; \log(c) = -5.6 \;\Rightarrow\; c = 10^{-5.6}.$

To approximate, split the exponent into a whole part and a decimal part:

$10^{-5.6} = 10^{-6+0.4} = 10^{-6}\cdot10^{0.4} \approx 10^{-6}(2.5) = 2.5\times10^{-6}.$

Write both concentrations straight from the definition of pH:

$c_{4} = 10^{-4}, \qquad c_{7} = 10^{-7}.$

Compare them as a single ratio, using the quotient rule for exponents, rather than approximating each one separately:

$\frac{c_{4}}{c_{7}} = \frac{10^{-4}}{10^{-7}} = 10^{-4-(-7)} = 10^{3}.$

The exponent in that ratio is exactly the pH gap, $7-4=3$, because pH is $-\log(c)$: every whole unit pH drops multiplies $c$ by another factor of $10$. So a pH gap of $3$ always produces the same factor between the two concentrations, whatever the two actual pH values happen to be.

Start from what $m$ and $n$ mean: $b^{m}=M$ and $b^{n}=N$. Divide the two exponential forms and apply the given exponent rule for a quotient of like bases:

$\frac{M}{N} = \frac{b^{m}}{b^{n}} = b^{m-n}.$

This equation says that raising $b$ to the exponent $m-n$ produces $M/N$. By the definition of a logarithm, the exponent that produces $M/N$ is exactly what $\log_b(M/N)$ means, so

$\log_{b}\!\left(\frac{M}{N}\right) = m - n = \log_{b}(M) - \log_{b}(N).$

Nothing in the argument used a specific value of $M$, $N$, or $b$: the chain of equalities holds for every allowed base and every pair of positive numbers, not only for an example anyone happens to have checked.

Compute the quotient first: $M/N = (-8)/(-2) = 4$, a positive number, so the left side $\log_b(4)$ is a perfectly good real number.

The right side needs $\log_b(M)$ and $\log_b(N)$ SEPARATELY, and neither input is positive: $M=-8$ and $N=-2$ are both negative, and the domain of a logarithm is $y>0$. So

$\log_{b}(M) \text{ and } \log_{b}(N) \text{ are both undefined}.$

An equation cannot hold between a real number and an expression that is not a number at all. Trace this back to part A: the very first line of that proof set $m=\log_b(M)$ and $n=\log_b(N)$, a step that only makes sense when $M$ and $N$ are positive. So the positivity assumption was not added for safety after the fact; it is what let the proof write down $m$ and $n$ in the first place.

Doubling means the balance reaches twice the deposit:

$2000(1.04)^{t} = 4000 \;\Rightarrow\; (1.04)^{t} = 2.$

Take the common log of both sides and use the power law to bring the exponent down:

$t\log(1.04) = \log(2) \;\Rightarrow\; t = \frac{\log(2)}{\log(1.04)} \approx \frac{0.3010}{0.0170} \approx 17.7.$

Set up the doubling equation for this deposit exactly as before:

$500(1.04)^{t} = 1000 \;\Rightarrow\; (1.04)^{t} = 2.$

The specific deposit has already divided out, leaving the identical equation solved in part A, so the same steps give

$t = \frac{\log(2)}{\log(1.04)} \approx 17.7 \text{ years}.$

Comparing the two results: the doubling time computed here matches the one from part A exactly, even though the two deposits are different.

Start from the general balance formula and set it equal to twice the deposit:

$P(1+r)^{t} = 2P.$

Divide both sides by $P$, which is a positive amount of money and so never $0$:

$(1+r)^{t} = 2.$

The deposit has left the equation entirely: it appeared on both sides in exactly the same way, so dividing removes it before a logarithm is ever taken. This holds for every positive $P$, not only the two amounts tried earlier, because nothing about the division used a particular value of $P$. The only quantity remaining in the equation is $r$, so the time it takes a balance to double is left depending on the interest rate alone.