Exponential functions and logarithms

Answer: $y = -6$

$2^{x - 3}$ gets as close to $0$ as you like as $x$ goes to $-\infty$ but never reaches it. Times $3$ it still tends to $0\mathclose{,}$ and then $- 6$ moves the asymptote: $y = -6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave $y = 0\mathclose{,}$ the asymptote of $2^{x}$ itself → $y = 0$
• Gave the $y$-intercept instead → $y = -\dfrac{45}{8}$

Answer: $\log_{2}(19)$

$19$ is not a whole power of $2\mathclose{,}$ so take the logarithm with base $2$ of both sides: $x = \log_{2}(19)\mathclose{.}$ (As a quotient of natural logarithms, $x = \dfrac{\ln 19}{\ln 2}\mathclose{.})$

Common mistakes, and the answer each one gives:

• Divided $19$ by $2$ → $\dfrac{19}{2}$
• Took $\ln 19$ and forgot to divide by $\ln 2$ → $\ln 19$

Answer: $8$

The product rule combines the left side: $\log_{2}\left(x(x - 7)\right) = 3\mathclose{,}$ so $x(x - 7) = 2^{3} = 8\mathclose{,}$ which is $x^{2} - 7 x - 8 = 0\mathclose{,}$ or $(x - 8)(x + 1) = 0\mathclose{.}$ Of $x = 8$ and $x = -1\mathclose{,}$ only $x = 8$ keeps both logarithms defined: $\log_{2}(-1)$ does not exist.

Common mistakes, and the answer each one gives:

• Kept $x = -1\mathclose{,}$ which makes the logarithms undefined → $-1$
• Added the arguments instead of multiplying them → $\frac{15}{2}$

Answer: $3$

At $x = 0\mathclose{,}$ $f(0) = a \cdot b^{0} = a\mathclose{,}$ so $a = 5\mathclose{.}$ Then $5 \cdot b^{3} = 135\mathclose{,}$ so $b^{3} = 27$ and, as $b > 0\mathclose{,}$ $b = 3\mathclose{:}$ the graph is multiplied by $3$ at every step of $1$ to the right.

Common mistakes, and the answer each one gives:

• Gave $b^3\mathclose{,}$ the growth over all $3$ steps → $27$
• Shared the growth out by dividing by $3$ instead of taking the root → $9$

Answer: $y = 3^{x + 2} + 2$

Shifting 2 units to the left replaces $x$ by $x + 2\mathclose{;}$ shifting 2 units up adds $2\mathclose{:}$ $y = 3^{x + 2} + 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Shifted the wrong way horizontally → $y = 3^{x - 2} + 2$
• Shifted the wrong way vertically → $y = 3^{x + 2} - 2$

Answer: $x = \log_{3}\left(\dfrac{y}{4}\right)$

First get the power on its own: $3^{x} = \dfrac{y}{4}\mathclose{.}$ Then take $\log_{3}$ of both sides: $x = \log_{3}\left(\dfrac{y}{4}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the logarithm before dividing by $4$ → $x = \dfrac{\log_{3}(y)}{4}$
• Subtracted $4$ instead of dividing by it → $x = \log_{3}\left(y - 4\right)$

Answer: $\frac{x^{6}}{6}$

The power rule turns $6\log_{5}(x)$ into $\log_{5}(x^{6})\mathclose{,}$ and the quotient rule combines the two into $\log_{5}\left(\frac{x^{6}}{6}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the arguments instead of dividing them → $x^{6} - 6$
• Divided the wrong way round → $\frac{6}{x^{6}}$
• Multiplied by the power instead of raising to it → $x$

Answer: $\dfrac{5}{2}$

Write both numbers as powers of $2\mathclose{:}$ $4 = 2^{2}$ and $32 = 2^{5}\mathclose{.}$ Then $\log_{4}(32) = \dfrac{5}{2}\mathclose{,}$ because $4$ raised to that power is $2^{5} = 32\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided the two numbers → $8$
• Turned the fraction of exponents upside down → $\frac{2}{5}$

Answer: $\dfrac{5}{4}$

Write the right-hand side as a power of $3\mathclose{:}$ $27 = 3^{3}\mathclose{.}$ With the same base on both sides the exponents must be equal, so $4 x - 2 = 3$ and $x = \dfrac{5}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Set the exponent equal to $27$ instead of to $3$ → $\frac{29}{4}$
• Forgot to move the constant in the exponent across → $\frac{3}{4}$

Answer: $8$

$\log_{2}(\text{something}) = 4$ means $\text{something} = 2^{4} = 16\mathclose{.}$ So $x + 8 = 16$ and $x = 8\mathclose{.}$ (Check: $x + 8$ is then positive, as a logarithm needs.)

Common mistakes, and the answer each one gives:

• Multiplied the base by $4$ instead of raising it to the power $4$ → $0$
• Forgot to subtract $8$ at the end → $16$

Answer: $x = -3$

A logarithm needs a positive argument: $x + 3 > 0\mathclose{,}$ so $x > -3\mathclose{.}$ As $x$ comes down to $-3$ the argument goes to $0$ and the logarithm to $-\infty\mathclose{:}$ the vertical asymptote is $x = -3\mathclose{.}$ The $- 3$ only moves the graph up or down.

Common mistakes, and the answer each one gives:

• Took the asymptote at $x = 3\mathclose{,}$ with the sign of the shift turned round → $x = 3$
• Gave where the graph crosses the $x$-axis instead → $x = 5$

Answer: $15$

On the $x$-axis $f(x) = 0\mathclose{:}$ $\log_{2}\left(x + 1\right) = 4\mathclose{,}$ so $x + 1 = 2^{4} = 16$ and $x = 15\mathclose{.}$

Common mistakes, and the answer each one gives:

• Solved $\log_{2}\left(x + 1\right) = -4\mathclose{,}$ with the sign of the 4 turned round → $-\dfrac{15}{16}$
• Multiplied the base by $4$ instead of raising it to the power $4$ → $7$

Answer: $1$

$9^{x} = \left(3^{x}\right)^{2}\mathclose{,}$ so put $u = 3^{x}\mathclose{:}$ $u^{2} + 2 u - 15 = (u - 3)(u + 5) = 0\mathclose{.}$ A power of $3$ is never negative, so $u = -5$ is impossible and $3^{x} = 3\mathclose{,}$ giving $x = 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave $u = 3^x = 3$ instead of $x$ → $3$

Answer: $x = \frac{1}{3} \text{ or } x = 9$

Put $u = \log_{3}(x)\mathclose{:}$ $u^{2} - u - 2 = (u + 1)(u - 2) = 0\mathclose{,}$ so $u = -1$ or $u = 2\mathclose{.}$ Then $x = 3^{u}\mathclose{:}$ $x = 3^{-1} = \frac{1}{3}$ or $x = 3^{2} = 9\mathclose{.}$ Both are positive, so both logarithms exist.

Common mistakes, and the answer each one gives:

• Gave the values of $u$ and stopped → $x = -1 \text{ or } x = 2$
• Took $x = 3u$ instead of $x = 3^u$ → $x = -3 \text{ or } x = 6$

Answer: $y = -\log_{2}\left(x - 4\right) - 2$

Reflecting in the $x$-axis changes the sign of $y\mathclose{:}$ $y = -\log_{2}(x)\mathclose{.}$ Shifting 4 units to the right replaces $x$ by $x - 4\mathclose{,}$ and 2 units down adds $-2\mathclose{:}$ $y = -\log_{2}\left(x - 4\right) - 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Reflected in the $y$-axis instead → $y = \log_{2}\left(4 - x\right) - 2$
• Shifted the wrong way horizontally → $y = -\log_{2}\left(x + 4\right) - 2$

Answer: $y = 2$

$2^{x + 1}$ gets as close to $0$ as you like as $x$ goes to $-\infty$ but never reaches it. Times $5$ it still tends to $0\mathclose{,}$ and then $+ 2$ moves the asymptote: $y = 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave $y = 0\mathclose{,}$ the asymptote of $2^{x}$ itself → $y = 0$
• Gave the $y$-intercept instead → $y = 12$

Answer: $\log_{2}(14)$

$14$ is not a whole power of $2\mathclose{,}$ so take the logarithm with base $2$ of both sides: $x = \log_{2}(14)\mathclose{.}$ (As a quotient of natural logarithms, $x = \dfrac{\ln 14}{\ln 2}\mathclose{.})$

Common mistakes, and the answer each one gives:

• Divided $14$ by $2$ → $7$
• Took $\ln 14$ and forgot to divide by $\ln 2$ → $\ln 14$

Answer: $4$

The product rule combines the left side: $\log_{2}\left(x(x - 2)\right) = 3\mathclose{,}$ so $x(x - 2) = 2^{3} = 8\mathclose{,}$ which is $x^{2} - 2 x - 8 = 0\mathclose{,}$ or $(x - 4)(x + 2) = 0\mathclose{.}$ Of $x = 4$ and $x = -2\mathclose{,}$ only $x = 4$ keeps both logarithms defined: $\log_{2}(-2)$ does not exist.

Common mistakes, and the answer each one gives:

• Kept $x = -2\mathclose{,}$ which makes the logarithms undefined → $-2$
• Added the arguments instead of multiplying them → $5$

Answer: $2$

At $x = 0\mathclose{,}$ $f(0) = a \cdot b^{0} = a\mathclose{,}$ so $a = 2\mathclose{.}$ Then $2 \cdot b^{3} = 16\mathclose{,}$ so $b^{3} = 8$ and, as $b > 0\mathclose{,}$ $b = 2\mathclose{:}$ the graph is multiplied by $2$ at every step of $1$ to the right.

Common mistakes, and the answer each one gives:

• Gave $b^3\mathclose{,}$ the growth over all $3$ steps → $8$
• Shared the growth out by dividing by $3$ instead of taking the root → $\dfrac{8}{3}$

Answer: $y = 3^{-x} + 3$

Reflecting in the $y$-axis replaces $x$ by $-x\mathclose{:}$ $y = 3^{-x}\mathclose{.}$ Shifting 3 units up adds $3\mathclose{:}$ $y = 3^{-x} + 3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Reflected in the $x$-axis instead → $y = -3^{x} + 3$
• Shifted the wrong way vertically → $y = 3^{-x} - 3$

Answer: $x = \log_{2}\left(\dfrac{y}{5}\right)$

First get the power on its own: $2^{x} = \dfrac{y}{5}\mathclose{.}$ Then take $\log_{2}$ of both sides: $x = \log_{2}\left(\dfrac{y}{5}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the logarithm before dividing by $5$ → $x = \dfrac{\log_{2}(y)}{5}$
• Subtracted $5$ instead of dividing by it → $x = \log_{2}\left(y - 5\right)$

Answer: $5 x^{7}$

The power rule turns $7\log_{6}(x)$ into $\log_{6}(x^{7})\mathclose{,}$ and the product rule combines the two into $\log_{6}\left(5 x^{7}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the arguments instead of multiplying them → $x^{7} + 5$
• Multiplied by the power instead of raising to it → $35 x$

Answer: $2$

Write both numbers as powers of $3\mathclose{:}$ $9 = 3^{2}$ and $81 = 3^{4}\mathclose{.}$ Then $\log_{9}(81) = \dfrac{4}{2} = 2\mathclose{,}$ because $9$ raised to that power is $3^{4} = 81\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided the two numbers → $9$
• Turned the fraction of exponents upside down → $\frac{1}{2}$

Answer: $-\dfrac{1}{4}$

Write the right-hand side as a power of $5\mathclose{:}$ $5 = 5^{1}\mathclose{.}$ With the same base on both sides the exponents must be equal, so $4 x + 2 = 1$ and $x = -\dfrac{1}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Set the exponent equal to $5$ instead of to $1$ → $\frac{3}{4}$
• Forgot to move the constant in the exponent across → $\frac{1}{4}$

Answer: $17$

$\log_{2}(\text{something}) = 3$ means $\text{something} = 2^{3} = 8\mathclose{.}$ So $x - 9 = 8$ and $x = 17\mathclose{.}$ (Check: $x - 9$ is then positive, as a logarithm needs.)

Common mistakes, and the answer each one gives:

• Multiplied the base by $3$ instead of raising it to the power $3$ → $15$
• Forgot to add $9$ at the end → $8$
• Raised the exponent to the power of the base, $3^{2}$ → $18$

Answer: $x = -2$

A logarithm needs a positive argument: $x + 2 > 0\mathclose{,}$ so $x > -2\mathclose{.}$ As $x$ comes down to $-2$ the argument goes to $0$ and the logarithm to $-\infty\mathclose{:}$ the vertical asymptote is $x = -2\mathclose{.}$ The $- 2$ only moves the graph up or down.

Common mistakes, and the answer each one gives:

• Took the asymptote at $x = 2\mathclose{,}$ with the sign of the shift turned round → $x = 2$
• Gave where the graph crosses the $x$-axis instead → $x = 7$

Answer: $12$

On the $x$-axis $f(x) = 0\mathclose{:}$ $\log_{3}\left(x - 3\right) = 2\mathclose{,}$ so $x - 3 = 3^{2} = 9$ and $x = 12\mathclose{.}$

Common mistakes, and the answer each one gives:

• Solved $\log_{3}\left(x - 3\right) = -2\mathclose{,}$ with the sign of the 2 turned round → $\dfrac{28}{9}$
• Multiplied the base by $2$ instead of raising it to the power $2$ → $9$

Answer: $3$

$4^{x} = \left(2^{x}\right)^{2}\mathclose{,}$ so put $u = 2^{x}\mathclose{:}$ $u^{2} - 4 u - 32 = (u - 8)(u + 4) = 0\mathclose{.}$ A power of $2$ is never negative, so $u = -4$ is impossible and $2^{x} = 8\mathclose{,}$ giving $x = 3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave $u = 2^x = 8$ instead of $x$ → $8$

Answer: $x = 3 \text{ or } x = 9$

Put $u = \log_{3}(x)\mathclose{:}$ $u^{2} - 3 u + 2 = (u - 1)(u - 2) = 0\mathclose{,}$ so $u = 1$ or $u = 2\mathclose{.}$ Then $x = 3^{u}\mathclose{:}$ $x = 3^{1} = 3$ or $x = 3^{2} = 9\mathclose{.}$ Both are positive, so both logarithms exist.

Common mistakes, and the answer each one gives:

• Gave the values of $u$ and stopped → $x = 1 \text{ or } x = 2$
• Took $x = 3u$ instead of $x = 3^u$ → $x = 3 \text{ or } x = 6$

Answer: $y = -\log_{3}\left(x + 2\right) - 3$

Reflecting in the $x$-axis changes the sign of $y\mathclose{:}$ $y = -\log_{3}(x)\mathclose{.}$ Shifting 2 units to the left replaces $x$ by $x + 2\mathclose{,}$ and 3 units down adds $-3\mathclose{:}$ $y = -\log_{3}\left(x + 2\right) - 3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Reflected in the $y$-axis instead → $y = \log_{3}\left(- x - 2\right) - 3$
• Shifted the wrong way horizontally → $y = -\log_{3}\left(x - 2\right) - 3$