Practice 1 Solutions

Expand.

$\log_{5} 7 x^{2}$

$\log_{5} 7 + \log_{5} x^{2}$

$\log_{5} 7 + 2 \log_{5} x$

$\log \frac{4 x}{(2 x + 1) (x - 2)}$

$\log 4 x - \log (2 x + 1) (x - 2) \log 4 x - [\log (2 x + 1) + \log (x - 2)]$

$\log 4 + \log x - \log (2 x + 1) - \log (x - 2)$

$\log_{x} \frac{x^{3} y^{5}}{a b^{2}}$

$\log_{x} x^{3} + \log_{x} y^{5} - (\log_{x} a + \log_{x} b^{2}) 3 \log_{x} x + \log_{x} y^{5} - \log_{x} a - \log_{x} b^{2} 3 + \log_{x} y^{5} - \log_{x} a - \log_{x} b^{2}$

$3 + 5 \log_{x} y - \log_{x} a - 2 \log_{x} b$

Note

Q: What is the value of $\log_{x} x$ ?

A: 1

$\log_{v} \frac{\sqrt[4]{x^{3} y}}{z^{5}}$

$\log_{v} \sqrt[4]{x^{3} y} - \log_{v} z^{5} \log_{v} x^{\frac{3}{4}} y^{\frac{1}{4}} - \log_{v} z^{5} \log_{v} x^{\frac{3}{4}} + \log_{v} y^{\frac{1}{4}} - \log_{v} z^{5}$

$\frac{3}{4} \log_{v} x + \frac{1}{4} \log_{v} y - 5 \log_{v} z$

Contract. Rewrite as a single log.

$3 \log_{x} (y + 2) - \log_{x} (y - 1)$

$\log_{x} {(y + 2)}^{3} - \log_{x} (y - 1)$

$\log_{x} \frac{{(y + 2)}^{3}}{(y - 1)}$

$2 \log_{7} x + 4 \log_{7} (x - 1) - 3 \log_{7} (x + 1)$

$\log_{7} x^{2} + \log_{7} {(x - 1)}^{4} - \log_{7} {(x + 1)}^{3}$

$\log_{7} \frac{x^{2} {(x - 1)}^{4}}{{(x + 1)}^{3}}$

$\frac{1}{4} \log x + \frac{3}{4} \log y - 2 \log z$

$\log x^{\frac{1}{4}} + \log y^{\frac{3}{4}} - \log z^{2} \log (x^{\frac{1}{4}} y^{\frac{3}{4}}) - \log z^{2} \log \frac{{(x^{1} y^{3})}^{\frac{1}{4}}}{z^{2}}$

$\log \frac{\sqrt[4]{x y^{3}}}{z^{2}}$

$\frac{2}{3} \log_{2} (x + 1) - 3 \log_{2} (2 x + 1)$

$\log_{2} {(x + 1)}^{\frac{2}{3}} - \log_{2} {(2 x + 1)}^{3} \log_{2} \frac{{({(x + 1)}^{2})}^{\frac{1}{3}}}{{(2 x + 1)}^{3}}$

$\log_{2} \frac{\sqrt[3]{{(x + 1)}^{2}}}{{(2 x + 1)}^{3}}$

Solve.

$\log_{3} (x - 4) + 5 = 7$

$\begin{array}{rcl} \log_{3} (x - 4) & = & 2 \\ 3^{2} & = & x - 4 \\ 9 & = & x - 4 \end{array}$

x = 13

$\log_{2} (x - 6) + \log_{2} (x - 4) - \log_{2} x = \log_{2} 4$

$\begin{array}{rcl} \log_{2} \frac{(x - 6) (x - 4)}{x} & = & \log_{2} 4 \\ \frac{(x - 6) (x - 4)}{x} & = & 4 \\ x^{2} - 10 x + 24 & = & 4 x \\ x^{2} - 14 x + 24 & = & 0 \\ (x - 2) (x - 12) & = & 0 \\ x & = & 2, 12 \end{array}$

x = 12

$\log_{2} 3 + \log_{2} x = 4$

$\begin{array}{rcl} \log_{2} 3 x & = & 4 \\ 2^{4} & = & 3 x \\ 16 & = & 3 x \end{array}$

$x = \frac{16}{3}$

$\log_{2} (x + 1) + \log_{2} (x - 1) = 3$

$\begin{array}{rcl} \log_{2} (x + 1) (x - 1) & = & 3 \\ \log_{2} (x^{2} - 1) & = & 3 \\ 2^{3} & = & x^{2} - 1 \\ 8 & = & x^{2} - 1 \\ x^{2} - 9 & = & 0 \\ (x + 3) (x - 3) & = & 0 \\ x & = & - 3, 3 \end{array}$

x = 3

Note

–3 is not a solution because the argument $(x - 1)$ must be greater than zero.

$\log (x + 4) = \log x + \log 4$

$\begin{array}{rcl} \log (x + 4) & = & \log 4 x \\ x + 4 & = & 4 x \\ 3 x & = & 4 \end{array}$

$x = \frac{4}{3}$

$\log (x + 2) - \log (4 x + 3) = \log (\frac{1}{x})$

$\begin{array}{rcl} \log (\frac{x + 2}{4 x + 3}) & = & \log (\frac{1}{x}) \\ \frac{x + 2}{4 x + 3} & = & \frac{1}{x} \\ x (x + 2) & = & 4 x + 3 \\ x^{2} + 2 x & = & 4 x + 3 \\ x^{2} - 2 x - 3 & = & 0 \\ (x + 1) (x - 3) & = & 0 \\ x & = & - 1, 3 \end{array}$

x = 3

$3 \log_{2} (4 x - 1) = 15$

$\begin{array}{rcl} \log_{2} (4 x - 1) & = & 5 \\ 2^{5} & = & 4 x - 1 \\ 32 & = & 4 x - 1 \\ 4 x & = & 33 \end{array}$

$x = \frac{33}{4}$

$\log_{6} (5 x + 1) = \log_{6} (2 x + 3) + \log_{6} 2$

$\begin{array}{rcl} \log_{6} (5 x + 1) & = & \log_{6} 2 (2 x + 3) \\ 5 x + 1 & = & 2 (2 x + 3) \\ 5 x + 1 & = & 4 x + 6 \end{array}$

x = 5

$\log_{2} (x + 2) - \log_{2} (x - 5) = 3$

$\begin{array}{rcl} \log_{2} (\frac{x + 2}{x - 5}) & = & 3 \\ 2^{3} & = & \frac{x + 2}{x - 5} \\ 8 (x - 5) & = & x + 2 \\ 8 x - 40 & = & x + 2 \\ 7 x & = & 42 \end{array}$

x = 6

$\log_{6} (x + 5) + \log_{6} x = 2$

$\begin{array}{rcl} \log_{6} [x (x + 5)] & = & 2 \\ 6^{2} & = & x (x + 5) \\ 36 & = & x^{2} + 5 x \\ x^{2} + 5 x - 36 & = & 0 \\ (x + 9) (x - 4) & = & 0 \\ x & = & - 9, 4 \end{array}$

x = 4

$\log_{5} (x - 2) - \log_{5} (x + 3) = \log_{5} (x + 1) - \log_{5} (x - 7)$

$\begin{array}{rcl} \log_{5} \frac{x - 2}{x + 3} & = & \log_{5} \frac{x + 1}{x - 7} \\ \frac{x - 2}{x + 3} & = & \frac{x + 1}{x - 7} \\ (x + 3) (x + 1) & = & (x - 2) (x - 7) \\ x^{2} + 4 x + 3 & = & x^{2} - 9 x + 14 \\ 4 x + 3 & = & - 9 x + 14 \\ 13 x & = & 11 \\ x & = & \frac{11}{13} \end{array}$

no solution

Note

When $\frac{11}{13}$ is substituted into the original equation, the result is the log of a negative number, which is undefined.

Q: What are the possible values of the argument?

A: The argument must be greater than zero.

$\log_{7} x + \log_{7} 4 = \log_{7} 2$

$\begin{array}{rcl} \log_{7} 4 x & = & \log_{7} 2 \\ 4 x & = & 2 \end{array}$

$x = \frac{1}{2}$