Practice 2 Solutions

Solve.

$64^{3 - x} = 32^{2 x + 1}$

$\begin{array}{rcl} 64 = 2^{6} 32 = 2^{5} \\ 2^{6 (3 - x)} & = & 2^{5 (2 x + 1)} \\ 6 (3 - x) & = & 5 (2 x + 1) \\ 18 - 6 x & = & 10 x + 5 \\ 16 x & = & 13 \end{array}$

$x = \frac{13}{16}$

$49^{a + 4} = 343^{a + 2}$

$\begin{array}{rcl} 49 = 7^{2} 343 = 7^{3} \\ 7^{2 (a + 4)} & = & 7^{3 (a + 2)} \\ 2 (a + 4) & = & 3 (a + 2) \\ 2 a + 8 & = & 3 a + 6 \end{array}$

$a = 2$

$625^{y} = {(\frac{1}{125})}^{- 2 y + 5}$

$\begin{array}{rcl} 625 & = & 5^{4} (\frac{1}{125}) = 5^{- 3} \\ 5^{4 (y)} & = & 5^{- 3 (- 2 y + 5)} \\ 4 y & = & - 3 (- 2 y + 5) \\ 4 y & = & 6 y - 15 \\ 2 y & = & 15 \end{array}$

$y = \frac{15}{2}$

${(\frac{1}{4})}^{3 x - 2} = 128^{x - 2}$

$\begin{array}{rcl} (\frac{1}{4}) = 2^{- 2} 128 = 2^{7} \\ 2^{- 2 (3 x - 2)} & = & 2^{7 (x)} \\ - 2 (3 x - 2) & = & 7 x \\ - 6 x + 4 & = & 7 x \\ 13 x & = & 4 \end{array}$

$x = \frac{4}{13}$

$81^{b + 5} = 243^{b}$

$\begin{array}{rcl} 81 = 3^{4} 243 = 3^{5} \\ 3^{4 (b + 5)} & = & 3^{5 (b)} \\ 4 (b + 5) & = & 5 b \\ 4 b + 20 & = & 5 b \end{array}$

$b = 20$

$36^{3 v - 4} = {(\frac{1}{216})}^{4 - v}$

$\begin{array}{rcl} 36 = 6^{2} (\frac{1}{216}) = 6^{- 3} \\ 6^{2 (3 v - 4)} & = & 6^{- 3 (4 - v)} \\ 2 (3 v - 4) & = & - 3 (4 - v) \\ 6 v - 8 & = & - 12 + 3 v \\ 3 v & = & - 4 \end{array}$

$v = - \frac{4}{3}$

$8^{x} = 64^{x}$

$64 = 8^{2} 8^{x} = 8^{2 (x)} x = 2 x$

$x = 0$

Note

Q: When x is substituted back into the equation, what will both sides equal?
A: 1, Recall $x^{0} = 1$

$27^{3 x} = {(\frac{1}{81})}^{2 - 3 x}$

$\begin{array}{rcl} 27 = 3^{3} (\frac{1}{81}) = 3^{- 4} \\ 3^{3 (3 x)} & = & 3^{- 4 (2 - 3 x)} \\ 3 (3 x) & = & - 4 (2 - 3 x) \\ 9 x & = & - 8 + 12 x \\ 3 x & = & 8 \end{array}$

$x = \frac{8}{3}$

$15^{3 n} \geq 225^{2 n + 1}$

$225 = 15^{2} 15^{3 n} \geq 15^{2 (2 n + 1)} 3 n \geq 2 (2 n + 1) 3 n \geq 4 n + 2 - n \geq 2$

$n \leq - 2$

$1000^{2 - 3 x} < 10000^{x + 4}$

$\begin{array}{rcl} 1000 = 10^{3} 10000 = 10^{4} \\ 10^{3 (2 - 3 x)} & < & 10^{4 (x + 4)} \\ 3 (2 - 3 x) & < & 4 (x + 4) \\ 6 - 9 x & < & 4 x + 16 \\ - 10 & < & 13 x \end{array}$

$x > - \frac{10}{13}$

${(\frac{1}{121})}^{3 h} \leq 11^{2 h - 3}$

$\begin{array}{rcl} (\frac{1}{121}) & = & 11^{- 2} \\ 11^{- 2 (3 h)} & \leq & 11^{2 h - 3} \\ - 2 (3 h) & \leq & 2 h - 3 \\ - 6 h & \leq & 2 h - 3 \\ - 8 h & \leq & - 3 \end{array}$

$h \geq \frac{3}{8}$

$8^{7 - 2 p} > 16^{5 - p}$

$\begin{array}{rcl} 8 = 2^{3} 16 = 2^{4} \\ 2^{3 (7 - 2 p)} & > & 2^{4 (5 - p)} \\ 3 (7 - 2 p) & > & 4 (5 - p) \\ 21 - 6 p & > & 20 - 4 p \\ 1 & > & 2 p \end{array}$

$p < \frac{1}{2}$

$81^{5 w + 2} < 27^{4 w + 2}$

$\begin{array}{rcl} 81 = 3^{4} 27 = 3^{3} \\ 3^{4 (5 w + 2)} & < & 3^{3 (4 w + 2)} \\ 4 (5 w + 2) & < & 3 (4 w + 2) \\ 20 w + 8 & < & 12 w + 6 \\ 8 w & < & - 2 \end{array}$

$w < - \frac{1}{4}$

$36^{2 k - 1} > 36^{- k}$

$2 k - 1 > - k 3 k > 1$

$k > \frac{1}{3}$