{"id":1830,"date":"2019-08-14T19:04:48","date_gmt":"2019-08-14T23:04:48","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=1830"},"modified":"2019-08-15T12:54:32","modified_gmt":"2019-08-15T16:54:32","slug":"answer-key-11-4","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/answer-key-11-4\/","title":{"raw":"Answer Key 11.4","rendered":"Answer Key 11.4"},"content":{"raw":"[latexpage]\r\n
    \r\n \t
  1. \\(\\begin{array}{rrrrrrrr}\r\n\\\\ \\\\\r\n&1&-&2n&=&1&-&3n \\\\\r\n-&1&+&3n&&-1&+&3n \\\\\r\n\\midrule\r\n&&&n&=&0&&\r\n\\end{array}\\)<\/li>\r\n \t
  2. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n4^{2x}&=&4^{-2} \\\\ \\\\\r\n\\dfrac{2x}{2}&=&\\dfrac{-2}{2} \\\\ \\\\\r\nx&=&-1\r\n\\end{array}\\)<\/li>\r\n \t
  3. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n4^{2a}&=&4^0 \\\\\r\n2a&=&0 \\\\\r\na&=&0\r\n\\end{array}\\)<\/li>\r\n \t
  4. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(4^2)^{-3p}&=&(4^3)^{3p} \\\\\r\n4^{-6p}&=&4^{9p} \\\\\r\n-6p&=&\\phantom{+}9p \\\\\r\n+6p&&+6p \\\\\r\n\\midrule\r\n0&=&15p \\\\ \\\\\r\np&=&0\r\n\\end{array}\\)<\/li>\r\n \t
  5. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(5^{-2})^{-k}&=&(5^3)^{-2k-2} \\\\\r\n5^{2k}&=&5^{-6k-6} \\\\\r\n2k&=&-6k-6 \\\\\r\n+6k&&+6k \\\\\r\n\\midrule\r\n8k&=&-6 \\\\ \\\\\r\nk&=&-\\dfrac{6}{8}\\Rightarrow -\\dfrac{3}{4}\r\n\\end{array}\\)<\/li>\r\n \t
  6. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(5^4)^{-n-2}&=&5^{-3} \\\\\r\n5^{-4n+8}&=&5^3 \\\\\r\n-4n+8&=&\\phantom{-}3 \\\\\r\n-8&&-8 \\\\\r\n\\midrule\r\n-4n&=&-5 \\\\ \\\\\r\nn&=&\\dfrac{5}{4}\r\n\\end{array}\\)<\/li>\r\n \t
  7. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n6^{2m+1}&=&6^{-2} \\\\\r\n2m+1&=&-2 \\\\\r\n-1&&-1 \\\\\r\n\\midrule\r\n2m&=&-3 \\\\ \\\\\r\nm&=&-\\dfrac{3}{2}\r\n\\end{array}\\)<\/li>\r\n \t
  8. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n2r-3&=&\\phantom{-}r-3 \\\\\r\n-r+3&&-r+3 \\\\\r\n\\midrule\r\nr&=&0\r\n\\end{array}\\)<\/li>\r\n \t
  9. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\\r\n6^{-3x}&=&6^2 \\\\\r\n\\therefore -3x&=&2 \\\\\r\nx&=&-\\dfrac{2}{3}\r\n\\end{array}\\)<\/li>\r\n \t
  10. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\n2n&=&-n \\\\\r\n+n&&+n \\\\\r\n\\midrule\r\n3n&=&0 \\\\ \\\\\r\nn&=&0\r\n\\end{array}\\)<\/li>\r\n \t
  11. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\\r\n(2^6)^b&=&2^5 \\\\\r\n6b&=&5 \\\\ \\\\\r\nb&=&\\dfrac{5}{6}\r\n\\end{array}\\)<\/li>\r\n \t
  12. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(6^3)^{-3v}&=&(6^2)^{3v} \\\\\r\n6^{-9v}&=&6^{6v} \\\\\r\n-9v&=&6v \\\\\r\n+9v&&+9v \\\\\r\n\\midrule\r\n0&=&15v \\\\ \\\\\r\nv&=&0\r\n\\end{array}\\)<\/li>\r\n \t
  13. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\\r\n(4^{-1})^x&=&4^2 \\\\\r\n4^{-x}&=&4^2 \\\\\r\n\\therefore -x&=&2 \\\\\r\nx&=&-2\r\n\\end{array}\\)<\/li>\r\n \t
  14. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(3^3)^{-2n-1}&=&3^2 \\\\\r\n3^{-6n-3}&=&3^2 \\\\\r\n-6n-3&=&2 \\\\\r\n+3&=&+3 \\\\\r\n\\midrule\r\n-6n&=&5 \\\\ \\\\\r\nn&=&-\\dfrac{5}{6}\r\n\\end{array}\\)<\/li>\r\n \t
  15. \\(\\begin{array}{rrl}\r\n\\\\\r\n\\therefore 3a&=&3 \\\\\r\na&=&1\r\n\\end{array}\\)<\/li>\r\n \t
  16. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n4^{-3v}&=&4^3 \\\\\r\n\\therefore -3v&=&3 \\\\\r\nv&=&-1\r\n\\end{array}\\)<\/li>\r\n \t
  17. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\n(6^2)^{3x}&=&\\phantom{-}(6^3)^{2x+1} \\\\\r\n6^{6x}&=&\\phantom{-}6^{6x+3} \\\\\r\n\\therefore 6x&=&\\phantom{-}6x+3 \\\\\r\n-6x&&-6x \\\\\r\n\\midrule\r\n0&=&3 \\Rightarrow \\text{no solution}\r\n\\end{array}\\)<\/li>\r\n \t
  18. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(4^3)^{x+2}&=&4^2 \\\\\r\n4^{3x+6}&=&4^2 \\\\\r\n\\therefore 3x+6 & =&\\phantom{-}2 \\\\\r\n-6&&-6 \\\\\r\n\\midrule\r\n3x&=&-4 \\\\ \\\\\r\nx&=&-\\dfrac{4}{3}\r\n\\end{array}\\)<\/li>\r\n \t
  19. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(3^2)^{2n+3}&=&3^5 \\\\\r\n3^{4n+6}&=&3^5 \\\\\r\n\\therefore 4n+6&=&\\phantom{-}5 \\\\\r\n-6&&-6 \\\\\r\n\\midrule\r\n4n&=&-1 \\\\ \\\\\r\nn&=&-\\dfrac{1}{4}\r\n\\end{array}\\)<\/li>\r\n \t
  20. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\n(4^2)^{2k}&=&4^{-3} \\\\\r\n4^{4k}&=&4^{-3} \\\\\r\n\\therefore 4k&=&-3 \\\\ \\\\\r\nk&=&-\\dfrac{3}{4}\r\n\\end{array}\\)<\/li>\r\n \t
  21. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n3x-2&=&\\phantom{-}3x+1 \\\\\r\n-3x+2&&-3x+2 \\\\\r\n\\midrule\r\n0&=&\\phantom{-}3\\Rightarrow \\text{no solution}\r\n\\end{array}\\)<\/li>\r\n \t
  22. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n(2^5)^p&=&(3^2)^{-3p} \\\\\r\n\\therefore 5p&=&-6p \\\\\r\n+6p&&+6p \\\\\r\n\\midrule\r\n11p&=&0 \\\\ \\\\\r\np&=&0\r\n\\end{array}\\)<\/li>\r\n \t
  23. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n-2x&=&3 \\\\\r\nx&=&-\\dfrac{3}{2}\r\n\\end{array}\\)<\/li>\r\n \t
  24. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n2n&=&2-3n \\\\\r\n+3n&&\\phantom{2}+3n \\\\\r\n\\midrule\r\n5n&=&2 \\\\ \\\\\r\nn&=&\\dfrac{2}{5}\r\n\\end{array}\\)<\/li>\r\n \t
  25. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\nm+2&=&-m \\\\\r\n+m-2&=&+m-2 \\\\\r\n\\midrule\r\n2m&=&-2 \\\\ \\\\\r\nm&=&-1\r\n\\end{array}\\)<\/li>\r\n \t
  26. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\n(5^4)^{2x}&=&5^2 \\\\\r\n5^{8x}&=&5^2 \\\\\r\n\\therefore 8x&=&2 \\\\ \\\\\r\nx&=&\\dfrac{1}{4}\r\n\\end{array}\\)<\/li>\r\n \t
  27. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(6^{-2})^{b-1}&=&6^3 \\\\\r\n6^{-2b+2}&=&6^3 \\\\\r\n\\therefore -2b+2&=&\\phantom{-}3 \\\\\r\n-2&& -2 \\\\\r\n\\midrule\r\n-2b&=&\\phantom{-}1 \\\\ \\\\\r\nb&=&-\\dfrac{1}{2}\r\n\\end{array}\\)<\/li>\r\n \t
  28. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\n(6^3)^{2n}&=&6^2 \\\\\r\n6^{6n}&=&6^2 \\\\\r\n\\therefore 6n&=&2 \\\\ \\\\\r\nn&=&\\dfrac{1}{3}\r\n\\end{array}\\)<\/li>\r\n \t
  29. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\n2-2x&=&\\phantom{-}2 \\\\\r\n-2\\phantom{-2x}&=&-2 \\\\\r\n\\midrule\r\n-2x&=&0 \\\\ \\\\\r\nx&=&0\r\n\\end{array}\\)<\/li>\r\n \t
  30. \\(\\begin{array}{rrl}\r\n\\\\ \\\\ \\\\ \\\\\r\n(2^{-2})^{3v-2}&=&\\phantom{-}(2^6)^{1-v} \\\\\r\n2^{-6v+4}&=&\\phantom{-}2^{6-6v} \\\\\r\n\\therefore -6v+4&=&\\phantom{-}6-6v \\\\\r\n+6v-4&& -4+6v \\\\\r\n\\midrule\r\n0&=&\\phantom{-}2\\Rightarrow \\text{No solution}\r\n\\end{array}\\)<\/li>\r\n<\/ol>","rendered":"
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