Answer Key 11.4

  1. \begin{array}{rrrrrrrr} \\ \\ &1&-&2n&=&1&-&3n \\ -&1&+&3n&&-1&+&3n \\ \midrule &&&n&=&0&& \end{array}
  2. \begin{array}{rrl} \\ \\ \\ \\ \\ 4^{2x}&=&4^{-2} \\ \\ \dfrac{2x}{2}&=&\dfrac{-2}{2} \\ \\ x&=&-1 \end{array}
  3. \begin{array}{rrl} \\ \\ 4^{2a}&=&4^0 \\ 2a&=&0 \\ a&=&0 \end{array}
  4. \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}
  5. \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}
  6. \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}
  7. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ 6^{2m+1}&=&6^{-2} \\ 2m+1&=&-2 \\ -1&&-1 \\ \midrule 2m&=&-3 \\ \\ m&=&-\dfrac{3}{2} \end{array}
  8. \begin{array}{rrl} \\ \\ 2r-3&=&\phantom{-}r-3 \\ -r+3&&-r+3 \\ \midrule r&=&0 \end{array}
  9. \begin{array}{rrl} \\ \\ \\ 6^{-3x}&=&6^2 \\ \therefore -3x&=&2 \\ x&=&-\dfrac{2}{3} \end{array}
  10. \begin{array}{rrl} \\ \\ \\ \\ 2n&=&-n \\ +n&&+n \\ \midrule 3n&=&0 \\ \\ n&=&0 \end{array}
  11. \begin{array}{rrl} \\ \\ \\ (2^6)^b&=&2^5 \\ 6b&=&5 \\ \\ b&=&\dfrac{5}{6} \end{array}
  12. \begin{array}{rrl} \\ \\ \\ \\ \\ \\ (6^3)^{-3v}&=&(6^2)^{3v} \\ 6^{-9v}&=&6^{6v} \\ -9v&=&6v \\ +9v&&+9v \\ \midrule 0&=&15v \\ \\ v&=&0 \end{array}
  13. \begin{array}{rrl} \\ \\ \\ (4^{-1})^x&=&4^2 \\ 4^{-x}&=&4^2 \\ \therefore -x&=&2 \\ x&=&-2 \end{array}
  14. \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}
  15. \begin{array}{rrl} \\ \therefore 3a&=&3 \\ a&=&1 \end{array}
  16. \begin{array}{rrl} \\ \\ 4^{-3v}&=&4^3 \\ \therefore -3v&=&3 \\ v&=&-1 \end{array}
  17. \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}
  18. \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}
  19. \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}
  20. \begin{array}{rrl} \\ \\ \\ \\ (4^2)^{2k}&=&4^{-3} \\ 4^{4k}&=&4^{-3} \\ \therefore 4k&=&-3 \\ \\ k&=&-\dfrac{3}{4} \end{array}
  21. \begin{array}{rrl} \\ \\ 3x-2&=&\phantom{-}3x+1 \\ -3x+2&&-3x+2 \\ \midrule 0&=&\phantom{-}3\Rightarrow \text{no solution} \end{array}
  22. \begin{array}{rrl} \\ \\ \\ \\ \\ (2^5)^p&=&(3^2)^{-3p} \\ \therefore 5p&=&-6p \\ +6p&&+6p \\ \midrule 11p&=&0 \\ \\ p&=&0 \end{array}
  23. \begin{array}{rrl} \\ \\ -2x&=&3 \\ x&=&-\dfrac{3}{2} \end{array}
  24. \begin{array}{rrl} \\ \\ \\ \\ \\ 2n&=&2-3n \\ +3n&&\phantom{2}+3n \\ \midrule 5n&=&2 \\ \\ n&=&\dfrac{2}{5} \end{array}
  25. \begin{array}{rrl} \\ \\ \\ \\ m+2&=&-m \\ +m-2&=&+m-2 \\ \midrule 2m&=&-2 \\ \\ m&=&-1 \end{array}
  26. \begin{array}{rrl} \\ \\ \\ \\ (5^4)^{2x}&=&5^2 \\ 5^{8x}&=&5^2 \\ \therefore 8x&=&2 \\ \\ x&=&\dfrac{1}{4} \end{array}
  27. \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}
  28. \begin{array}{rrl} \\ \\ \\ \\ (6^3)^{2n}&=&6^2 \\ 6^{6n}&=&6^2 \\ \therefore 6n&=&2 \\ \\ n&=&\dfrac{1}{3} \end{array}
  29. \begin{array}{rrl} \\ \\ \\ \\ 2-2x&=&\phantom{-}2 \\ -2\phantom{-2x}&=&-2 \\ \midrule -2x&=&0 \\ \\ x&=&0 \end{array}
  30. \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}

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Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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