Logarithms

$b^y = x$ is the same as $\log_b x = y$. Here $2^5 = 32$ becomes $\log_2 32 = 5$ (base stays the base, exponent becomes the log). ($\log_5 32 = 2$ swapped the base and exponent.)

$\log_3 81$ asks "3 to what power gives 81?" Since $3^4 = 81$, the answer is 4. (27 is $3^3$, not the index.)

A difference of $2$ in magnitude means $\log_{10}\dfrac{I_6}{I_4} = 6 - 4 = 2$, so $\dfrac{I_6}{I_4} = 10^2 = 100$. The magnitude 6 quake is $100$ times more intense.

$\text{pH} = -\log_{10}(10^{-3}) = -(-3) = 3$. (The negative sign in the formula turns the negative index into a positive pH.)

$\tfrac{1}{8} = 2^{-3}$, so $\log_2 \tfrac{1}{8} = -3$ (a number below 1 gives a negative log for base $> 1$). (3 ignored the reciprocal.)

The product law: $\log m + \log n = \log(mn)$, so $\log 6 + \log 5 = \log 30$. ($\log 11$ added the numbers instead of multiplying them.)

The quotient law: $\log_2 12 - \log_2 3 = \log_2 \tfrac{12}{3} = \log_2 4 = 2$. ($\log_2 9$ subtracted the numbers; $\log_2 36$ multiplied them.)

The power law brings the exponent to the front: $\log_b(x^p) = p\log_b x$, so $\log_b(x^3) = 3\log_b x$. ($(\log_b x)^3$ cubes the whole logarithm, which is different.)

Rewrite in index form: $x = 2^5 = 32$. (10 used $2 \times 5$; 25 used $5^2$.)

Index form: $x^4 = 81$, so $x = \sqrt[4]{81} = 3$ (since $3^4 = 81$). (9 solves $x^2 = 81$, not $x^4$.)

A note $n$ octaves up has frequency $440 \times 2^n$. Setting $440 \times 2^n = 3520$ gives $2^n = 8$, so $n = \log_2 8 = 3$ octaves.

Write both in base 2: $4^{3/2} = (2^2)^{3/2} = 2^3 = 8$, so $\log_4 8 = \tfrac{3}{2}$. (The answer need not be a whole number when the base is not a factor.)

$6 = 2 \times 3$, so $\log 6 = \log 2 + \log 3 \approx 0.301 + 0.477 = 0.778$. ($0.143$ multiplied the two values instead of adding.)

A logarithm is the inverse of an exponential with the same base, and a function and its inverse are always reflections of each other in the line $y = x$.

Take logs of both sides: $x\log 2 = \log 20$, so $x = \dfrac{\log 20}{\log 2} \approx \dfrac{1.301}{0.301} \approx 4.32$. (Since $2^4 = 16$ and $2^5 = 32$, the answer must lie between 4 and 5.)

Combine using the product law: $\log_3[x(x + 6)] = 3$, so $x(x + 6) = 3^3 = 27$. Then $x^2 + 6x - 27 = 0$ and $(x + 9)(x - 3) = 0$, giving $x = 3$ or $x = -9$. But $x = -9$ makes $\log_3 x$ undefined, so $x = 3$ only.

$\log_3 27 = 3$ (since $3^3 = 27$) and $\log_2 16 = 4$ (since $2^4 = 16$), so the sum is $3 + 4 = 7$. (These are separate logs with different bases, so they cannot be combined into one log.)

The product law $\log_b(mn) = \log_b m + \log_b n$ is a genuine identity. The others are classic traps: a log of a sum is not the sum of logs, and the power law gives $\log_b(m^n) = n\log_b m$, not $(\log_b m)^n$.

$\log_2(-8)$ asks for a power $y$ with $2^y = -8$. But $2^y$ is positive for every real $y$, so no such power exists. Logarithms are only defined for positive numbers.

Combine: $\log_2[x(x - 2)] = 3$, so $x(x - 2) = 2^3 = 8$, giving $x^2 - 2x - 8 = 0$ and $(x - 4)(x + 2) = 0$. So $x = 4$ or $x = -2$, but $x = -2$ makes $\log_2 x$ undefined, so $x = 4$ only.

The power law gives $2\log x = \log x^2$, then the quotient law: $\log x^2 - \log y = \log \dfrac{x^2}{y}$. ($\log \dfrac{2x}{y}$ treated $2\log x$ as $\log 2x$.)

$\log_2 1 = 0$, so $(1, 0)$ is on the curve (every log graph cuts the x-axis at $x = 1$). The point $(0, 1)$ cannot lie on it because $\log_2 0$ is undefined; $(8, 2)$ is wrong since $\log_2 8 = 3$.

The change-of-base rule gives $\log_5 20 = \dfrac{\log 20}{\log 5} \approx \dfrac{1.301}{0.699} \approx 1.86$. (Since $5^1 = 5$ and $5^2 = 25$, the answer must lie between 1 and 2.)

$40 = 8 \times 5 = 2^3 \times 5$, so $\log_2 40 = \log_2 2^3 + \log_2 5 = 3 + k$. ($8k$ multiplied instead of using the product law.)

$\dfrac{I}{I_0} = 1000$, so $L = 10\log_{10} 1000 = 10 \times 3 = 30$ decibels. (3 forgot to multiply by 10; 1000 forgot to take the logarithm.)