Logarithmic Functions

Approach:

The argument of a logarithm $y = \log_a x$ with $a > 0$ and $a \neq 1$ must be positive, so the domain is $(0, +\infty)$; since $a^x$ runs through all positive numbers on $\mathbb{R}$, its inverse takes every real value.

Steps:

1. For the logarithmic expression $\log_a x$ to be meaningful the argument must satisfy $x > 0$, so the domain is $(0, +\infty)$.
2. $y = \log_a x$ is the inverse of the exponential function $y = a^x$, whose range is $(0, +\infty)$, so the range of the inverse is $\mathbb{R}$.
3. As $x \to 0^+$ we have $y \to -\infty$ and as $x \to +\infty$ we have $y \to +\infty$, so the graph covers every real value on its domain.

Answer:

Domain: $(0, +\infty)$; Range: $\mathbb{R}$.

Approach:

The criterion for a logarithmic function is that it can be written as $y = \log_a x$ with $a > 0$ and $a \neq 1$, that is, the argument sits inside the logarithm and the base is a constant; match each of the four functions against that form.

Steps:

1. $y = \log_3 x$ is a logarithmic function with base $3 > 0$ and $3 \neq 1$.
2. $y = x^3$ is a power function, so it is neither a logarithmic nor an exponential function.
3. $y = \ln x$ is a logarithmic function with the natural constant $e$ as base, that is $\ln x = \log_e x$.
4. $y = 2^x$ is an exponential function; the argument $x$ does not sit inside a logarithm, so it is not a logarithmic function.

Answer:

$y = \log_3 x$, $y = \ln x$ are logarithmic functions, $y = x^3$, $y = 2^x$ are not.

Approach:

Every logarithmic graph passes through the fixed point $(1, 0)$, and its direction of monotonicity is decided solely by how the base compares with $1$; that fixes the shape of both curves and allows the comparison.

Steps:

1. The graph of a logarithmic function always passes through $(1, 0)$ and has the $y$-axis as an asymptote.
2. For $y = \log_2 x$ the base $2 > 1$, so the function is monotonically increasing on $(0, +\infty)$; the graph rises along the $y$-axis from the lower left.
3. For $y = \log_{1/2} x$ the base $0 < 1/2 < 1$, so the function is monotonically decreasing on $(0, +\infty)$; the graph falls along the $y$-axis from the upper left.
4. Since $\log_{1/2} x = \frac{\ln x}{\ln (1/2)} = -\log_2 x$, the two graphs are symmetric about the $x$-axis and their directions of monotonicity are opposite.

Answer:

$y = \log_2 x$ is monotonically increasing, $y = \log_{1/2} x$ is monotonically decreasing.

Approach:

An asymptote is a line the graph approaches arbitrarily closely without ever meeting it; for $y = \log_a x$ with $a > 1$ it is enough to study the behaviour of the function as $x$ tends to $0^+$.

Steps:

1. From $\log_a x = \frac{\ln x}{\ln a}$ and $\ln a > 0$ for $a > 1$, we see that $\ln x \to -\infty$ and hence $y \to -\infty$ as $x \to 0^+$.
2. To the left of the $y$-axis the graph therefore drops without bound as $x \to 0^+$ and hugs the $y$-axis.
3. Since the domain is $(0, +\infty)$, the graph never meets the $y$-axis, so the $y$-axis, that is $x = 0$, is the vertical asymptote.

Answer:

The y-axis ($x=0$) is its asymptote.

Approach:

To decide whether a function is logarithmic, check only whether it can be written as $y = \log_a x$ with a base satisfying $a > 0$ and $a \neq 1$.

Steps:

1. The base of $y = \log_{-2} x$ is $-2$, which does not satisfy the requirement $a > 0$.
2. An arbitrary real power of a negative base cannot be defined for every positive argument, for instance $(-2)^{1/2} = \sqrt{-2}$ is not a real number.
3. Since the base of a logarithmic function must be positive, $\log_{-2} x$ cannot be treated as a logarithmic function.

Answer:

No. The base $a$ of a logarithmic function must be greater than 0 and not equal to 1.

Approach:

Use the change of base formula to rewrite the logarithm with base $4$ in terms of natural logarithms so that approximate values can be used, then verify the result directly from the definition.

Steps:

1. The change of base formula gives $\log_4 16 = \frac{\ln 16}{\ln 4}$.
2. Substituting the values, $\ln 16 \approx 2.7726$ and $\ln 4 \approx 1.3863$.
3. Dividing gives $\frac{2.7726}{1.3863} = 2$.
4. Checking directly from the definition, $4^2 = 16$, so $\log_4 16 = 2$, in agreement with the change of base computation.

Answer:

Using the change of base formula: $\log_4 16 = \frac{\ln 16}{\ln 4} = \frac{2.7726}{1.3863} = 2$.

Or directly calculate: $4^2 = 16$, so $\log_4 16 = 2$.

Approach:

Use the change of base formula to rewrite the logarithm with base $3$ in terms of natural logarithms, then verify the result directly from the definition.

Steps:

1. The change of base formula gives $\log_3 27 = \frac{\ln 27}{\ln 3}$.
2. Substituting the values, $\ln 27 \approx 3.2958$ and $\ln 3 \approx 1.0986$.
3. Dividing gives $\frac{3.2958}{1.0986} = 3$.
4. Checking directly from the definition, $3^3 = 27$, so $\log_3 27 = 3$, in agreement with the change of base computation.

Answer:

Using the change of base formula: $\log_3 27 = \frac{\ln 27}{\ln 3} = \frac{3.2958}{1.0986} = 3$.

Or directly calculate: $3^3 = 27$, so $\log_3 27 = 3$.

Approach:

Rewrite both sides with natural logarithms through the change of base formula, multiply the two fractions and watch the intermediate factor $\ln b$ cancel out.

Steps:

1. By the change of base formula, $\log_a b = \frac{\ln b}{\ln a}$.
2. Likewise, $\log_b c = \frac{\ln c}{\ln b}$.
3. Multiplying, $\log_a b \cdot \log_b c = \frac{\ln b}{\ln a} \cdot \frac{\ln c}{\ln b} = \frac{\ln c}{\ln a}$, where the non-zero factor $\ln b$ cancels.
4. Applying the change of base formula once more, $\frac{\ln c}{\ln a} = \log_a c$, which proves the identity.

Answer:

Using the change of base formula: $\log_a b = \frac{\ln b}{\ln a}$, $\log_b c = \frac{\ln c}{\ln b}$

Therefore: $\log_a b \cdot \log_b c = \frac{\ln b}{\ln a} \cdot \frac{\ln c}{\ln b} = \frac{\ln c}{\ln a} = \log_a c$

Approach:

The two given function values are direct consequences of the logarithmic identities, so they impose no extra restriction on the base; verify each of them and then check the allowed range of the base.

Steps:

1. For every base with $a > 0$ and $a \neq 1$ we have $\log_a 1 = 0$, so $y(1) = 0$ always holds.
2. Likewise $\log_a a = 1$, so $y(a) = 1$ always holds as well.
3. Neither condition adds a new constraint, so the only remaining restriction is the given $a > 1$, which already guarantees a positive base different from $1$.

Answer:

Since $y(1) = \log_a 1 = 0$ is always true, and $y(a) = \log_a a = 1$ is also always true; therefore, as long as $a > 1$, $y = \log_a x$ satisfies the conditions, and $a$ can be any value greater than 1.

Approach:

Check for each option whether it can be written as $y = \log_a x$ with $a > 0$ and $a \neq 1$, that is, whether the argument appears inside a logarithm.

Steps:

1. (A) $y = \log_2 x$ has base $2 > 0$ with $2 \neq 1$, so it is a logarithmic function.
2. (B) $y = x^2$ is a power function and its argument does not appear inside a logarithm, so it is not a logarithmic function.
3. (C) $y = \ln x = \log_e x$ has base $e > 0$ with $e \neq 1$, so it is a logarithmic function.
4. (D) $y = 2^x$ is an exponential function and does not have the form of a logarithmic function.

Answer:

(A), (C) are logarithmic functions, (B), (D) are not.

Approach:

The monotonicity of a logarithmic function is decided solely by how the base compares with $1$: a base above $1$ gives an increasing function, a base between $0$ and $1$ gives a decreasing one.

Steps:

1. For any $x_1 < x_2$, since $0 < a < 1$ the exponential function $a^x$ is monotonically decreasing on $\mathbb{R}$, so $a^{x_1} > a^{x_2}$.
2. A larger argument therefore gives a smaller logarithmic value, so $y = \log_a x$ is monotonically decreasing on $(0, +\infty)$.
3. Accordingly the graph passes through $(1, 0)$ and falls along the $y$-axis from the upper left, mirroring the case $a > 1$ about the $x$-axis.

Answer:

$y$ is monotonically decreasing.

Approach:

Rewrite $\log_6 15$ with base $2$, so that the given values $\log_2 3$ and $\log_2 5$ appear directly, and then split products and quotients with the logarithm laws.

Steps:

1. The change of base formula gives $\log_6 15 = \frac{\log_2 15}{\log_2 6}$.
2. Factor the argument and the base separately: $15 = 3 \times 5$ and $6 = 2 \times 3$.
3. Expand with the product and quotient laws of logarithms: $\frac{\log_2 3 + \log_2 5}{\log_2 2 + \log_2 3}$.
4. Substituting $\log_2 3 = a$, $\log_2 5 = b$ and $\log_2 2 = 1$ gives $\frac{a + b}{1 + a}$.

Answer:

Using the change of base formula: $\log_6 15 = \frac{\log_2 15}{\log_2 6} = \frac{\log_2 (3 \times 5)}{\log_2 (2 \times 3)} = \frac{\log_2 3 + \log_2 5}{\log_2 2 + \log_2 3} = \frac{a + b}{1 + a}$