Practice 1 Solutions

Solve.

Remember, there must be a common base before solving for an indicated variable.

${(\frac{2}{5})}^{3 x + 1} = (\frac{8}{125})$

$\frac{8}{125} = {(\frac{2}{5})}^{3} \begin{matrix} {(\frac{2}{5})}^{3 x + 1} = {(\frac{2}{5})}^{3} \\ 3 x + 1 = 3 \\ 3 x = 2 \end{matrix}$

$x = \frac{2}{3}$

${(\frac{1}{27})}^{2 d} = 81^{d - 5}$

$\begin{array}{rcl} \begin{matrix} (\frac{1}{27}) = 3^{- 3} & 81 = 3^{4} \\ 3^{- 3 (2 d)} = 3^{4 (d - 5)} - 3 (2 d) = 4 (d - 5) - 6 d = 4 d - 20 - 10 d = - 20 \end{matrix} \end{array}$

$d = 2$

Note

Q: Why is it not appropriate to write 81 as $9^{2}$ for this problem?

A: Because 9 is not a common base for 27.

$36^{4 c - 3} = 216^{c + 1}$

$\begin{array}{rcl} 36 = 6^{2} 216 = 6^{3} \\ 6^{2 (4 c - 3)} & = & 6^{3 (c + 1)} \\ 2 (4 c - 3) & = & 3 (c + 1) \\ 8 c - 6 & = & 3 c + 3 \\ 5 c & = & 9 \end{array}$

$c = \frac{9}{5}$

$8^{2 y - 5} = 16^{y}$

$\begin{array}{rcl} 8 = 2^{3} 16 = 2^{4} \\ 2^{3 (2 y - 5)} & = & 2^{4 (y)} \\ 3 (2 y - 5) & = & 4 y \\ 6 y - 15 & = & 4 y \\ 2 y & = & 15 \end{array}$

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

Note

Remember to use a common base, not multiplication facts. A common error is writing $16 = 8 (2)$ , which is true, but not useful for solving this problem.

$100^{2 v} = 1000^{v + 3}$

$\begin{array}{rcl} 100 = 10^{2} 1000 = 10^{3} \\ 10^{2 (2 v)} & = & 10^{3 (v + 3)} \\ 2 (2 v) & = & 3 (v + 3) \\ 4 v & = & 3 v + 9 \end{array}$

$v = 9$

${(\frac{1}{8})}^{u + 3} = {(\frac{1}{32})}^{u}$

$\begin{array}{rcl} (\frac{1}{8}) = 2^{- 3} (\frac{1}{32}) = 2^{- 4} \\ 2^{- 3 (u + 3)} & = & 2^{- 4 (u)} \\ - 3 (u + 3) & = & - 4 (u) \\ - 3 u - 9 & = & - 4 u \end{array}$

$u = 9$

${(\frac{25}{81})}^{2 m + 1} = {(\frac{729}{125})}^{- 3 m}$

$\begin{array}{rcl} \frac{25}{81} = {(\frac{5}{9})}^{2} (\frac{729}{125}) = {(\frac{5}{9})}^{- 3} \\ {(\frac{5}{9})}^{2 (2 m + 1)} & = & {(\frac{5}{9})}^{- 3 (- 3 m)} \\ 2 (2 m + 1) & = & - 3 (- 3 m) \\ 4 m + 2 & = & 9 m \\ 5 m & = & 2 \end{array}$

$m = \frac{2}{5}$

Note

You can also solve this problem by taking the reciprocal of the first fraction. The solution remains the same.

$7^{5 x} = 343^{x - 2}$

$\begin{array}{rcl} 343 & = & 7^{3} \\ 7^{5 x} & = & 7^{3 (x - 2)} \\ 5 x & = & 3 (x - 2) \\ 5 x & = & 3 x - 6 \\ 2 x & = & - 6 \end{array}$

$x = - 3$

$25^{2 t - 6} = 125^{t}$

$\begin{array}{rcl} 25 = 5^{2} 125 = 5^{3} \\ 5^{2 (2 t - 6)} & = & 5^{3 (t)} \\ 2 (2 t - 6) & = & 3 t \\ 4 t - 12 & = & 3 t \\ - 12 & = & - t \end{array}$

t = 12

${(\frac{1}{36})}^{p + 3} < 216^{- 2 p}$

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

$p < \frac{3}{2}$

${(\frac{4}{9})}^{3 - x} > {(\frac{8}{27})}^{x + 1}$

$\begin{array}{rcl} \frac{4}{9} = {(\frac{2}{3})}^{2} \frac{8}{27} = {(\frac{2}{3})}^{3} \\ {(\frac{2}{3})}^{2 (3 - x)} & > & {(\frac{2}{3})}^{3 (x + 1)} \\ 2 (3 - x) & > & 3 (x + 1) \\ 6 - 2 x & > & 3 x + 3 \\ 3 & > & 5 x \end{array}$

$x < \frac{3}{5}$

$9^{- x + 5} \geq 27^{\frac{1}{3} x + 2}$

$\begin{array}{rcl} 9 = 3^{2} 27 = 3^{3} \\ 3^{2 (- x + 5)} & \geq & 3^{3 (\frac{1}{3} x + 2)} \\ 2 (- x + 5) & \geq & 3 (\frac{1}{3} x + 2) \\ - 2 x + 10 & \geq & x + 6 \\ - 3 x & \geq & - 4 \end{array}$

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

${(\frac{1}{4})}^{- z + 8} < {(\frac{1}{8})}^{2 z}$

$\begin{array}{rcl} \frac{1}{4} = 2^{- 2} \frac{1}{8} = 2^{- 3} \\ 2^{- 2 (- z + 8)} & < & 2^{- 3 (2 z)} \\ - 2 (- z + 8) & < & - 3 (2 z) \\ 2 z - 16 & < & - 6 z \\ 8 z & < & 16 \end{array}$

$z < 2$

$25^{3 x + 7} > 25^{2 x + 1}$

$3 x + 7 > 2 x + 1$

$x > - 6$

Note

The bases are both 25. You could change each base to 5 and get the same answer.