Answer Key 11.4
[latexpage]
- \(\begin{array}{rrrrrrrr}
\\ \\
&1&-&2n&=&1&-&3n \\
-&1&+&3n&&-1&+&3n \\
\midrule
&&&n&=&0&&
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\
4^{2x}&=&4^{-2} \\ \\
\dfrac{2x}{2}&=&\dfrac{-2}{2} \\ \\
x&=&-1
\end{array}\) - \(\begin{array}{rrl}
\\ \\
4^{2a}&=&4^0 \\
2a&=&0 \\
a&=&0
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\
(4^2)^{-3p}&=&(4^3)^{3p} \\
4^{-6p}&=&4^{9p} \\
-6p&=&\phantom{+}9p \\
+6p&&+6p \\
\midrule
0&=&15p \\ \\
p&=&0
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\
(5^{-2})^{-k}&=&(5^3)^{-2k-2} \\
5^{2k}&=&5^{-6k-6} \\
2k&=&-6k-6 \\
+6k&&+6k \\
\midrule
8k&=&-6 \\ \\
k&=&-\dfrac{6}{8}\Rightarrow -\dfrac{3}{4}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\
(5^4)^{-n-2}&=&5^{-3} \\
5^{-4n+8}&=&5^3 \\
-4n+8&=&\phantom{-}3 \\
-8&&-8 \\
\midrule
-4n&=&-5 \\ \\
n&=&\dfrac{5}{4}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\
6^{2m+1}&=&6^{-2} \\
2m+1&=&-2 \\
-1&&-1 \\
\midrule
2m&=&-3 \\ \\
m&=&-\dfrac{3}{2}
\end{array}\) - \(\begin{array}{rrl}
\\ \\
2r-3&=&\phantom{-}r-3 \\
-r+3&&-r+3 \\
\midrule
r&=&0
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\
6^{-3x}&=&6^2 \\
\therefore -3x&=&2 \\
x&=&-\dfrac{2}{3}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\
2n&=&-n \\
+n&&+n \\
\midrule
3n&=&0 \\ \\
n&=&0
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\
(2^6)^b&=&2^5 \\
6b&=&5 \\ \\
b&=&\dfrac{5}{6}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\
(6^3)^{-3v}&=&(6^2)^{3v} \\
6^{-9v}&=&6^{6v} \\
-9v&=&6v \\
+9v&&+9v \\
\midrule
0&=&15v \\ \\
v&=&0
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\
(4^{-1})^x&=&4^2 \\
4^{-x}&=&4^2 \\
\therefore -x&=&2 \\
x&=&-2
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\
(3^3)^{-2n-1}&=&3^2 \\
3^{-6n-3}&=&3^2 \\
-6n-3&=&2 \\
+3&=&+3 \\
\midrule
-6n&=&5 \\ \\
n&=&-\dfrac{5}{6}
\end{array}\) - \(\begin{array}{rrl}
\\
\therefore 3a&=&3 \\
a&=&1
\end{array}\) - \(\begin{array}{rrl}
\\ \\
4^{-3v}&=&4^3 \\
\therefore -3v&=&3 \\
v&=&-1
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\
(6^2)^{3x}&=&\phantom{-}(6^3)^{2x+1} \\
6^{6x}&=&\phantom{-}6^{6x+3} \\
\therefore 6x&=&\phantom{-}6x+3 \\
-6x&&-6x \\
\midrule
0&=&3 \Rightarrow \text{no solution}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\
(4^3)^{x+2}&=&4^2 \\
4^{3x+6}&=&4^2 \\
\therefore 3x+6 & =&\phantom{-}2 \\
-6&&-6 \\
\midrule
3x&=&-4 \\ \\
x&=&-\dfrac{4}{3}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\
(3^2)^{2n+3}&=&3^5 \\
3^{4n+6}&=&3^5 \\
\therefore 4n+6&=&\phantom{-}5 \\
-6&&-6 \\
\midrule
4n&=&-1 \\ \\
n&=&-\dfrac{1}{4}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\
(4^2)^{2k}&=&4^{-3} \\
4^{4k}&=&4^{-3} \\
\therefore 4k&=&-3 \\ \\
k&=&-\dfrac{3}{4}
\end{array}\) - \(\begin{array}{rrl}
\\ \\
3x-2&=&\phantom{-}3x+1 \\
-3x+2&&-3x+2 \\
\midrule
0&=&\phantom{-}3\Rightarrow \text{no solution}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\
(2^5)^p&=&(3^2)^{-3p} \\
\therefore 5p&=&-6p \\
+6p&&+6p \\
\midrule
11p&=&0 \\ \\
p&=&0
\end{array}\) - \(\begin{array}{rrl}
\\ \\
-2x&=&3 \\
x&=&-\dfrac{3}{2}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\
2n&=&2-3n \\
+3n&&\phantom{2}+3n \\
\midrule
5n&=&2 \\ \\
n&=&\dfrac{2}{5}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\
m+2&=&-m \\
+m-2&=&+m-2 \\
\midrule
2m&=&-2 \\ \\
m&=&-1
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\
(5^4)^{2x}&=&5^2 \\
5^{8x}&=&5^2 \\
\therefore 8x&=&2 \\ \\
x&=&\dfrac{1}{4}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\ \\ \\ \\
(6^{-2})^{b-1}&=&6^3 \\
6^{-2b+2}&=&6^3 \\
\therefore -2b+2&=&\phantom{-}3 \\
-2&& -2 \\
\midrule
-2b&=&\phantom{-}1 \\ \\
b&=&-\dfrac{1}{2}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\
(6^3)^{2n}&=&6^2 \\
6^{6n}&=&6^2 \\
\therefore 6n&=&2 \\ \\
n&=&\dfrac{1}{3}
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\
2-2x&=&\phantom{-}2 \\
-2\phantom{-2x}&=&-2 \\
\midrule
-2x&=&0 \\ \\
x&=&0
\end{array}\) - \(\begin{array}{rrl}
\\ \\ \\ \\
(2^{-2})^{3v-2}&=&\phantom{-}(2^6)^{1-v} \\
2^{-6v+4}&=&\phantom{-}2^{6-6v} \\
\therefore -6v+4&=&\phantom{-}6-6v \\
+6v-4&& -4+6v \\
\midrule
0&=&\phantom{-}2\Rightarrow \text{No solution}
\end{array}\)
