IAT SOLUTIONS - C_7.pdf

Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.4. 3 x 1 1 2 5 5 103 x 1 1 5 15x 1 1 5 log 315log 15x 5 }log 3 2 1x ø 1.46Check: 3 x 1 1 2 5 5 103 1.46 1 1 2 5 0 103 2.46 2 5 0 1014.92 2 5 0 1010 0 10 ✓5. log 4(4x 1 7) 5 log 411x4x 1 7 5 11x7 5 7x1 5 xCheck: log 4(4x 1 7) 5 log 411xlog 4(4(1) 1 7) 0 log 411(1)6. ln (3x 2 2) 5 ln 6x3x 2 2 5 6x22 5 3x2 2 }3 5 xlog 411 5 log 411 ✓Check: ln (3x 2 2) 5 ln 6xln 1312 }2 32 2 22 0 ln 612 }2 32ln (24) 5 ln (24) ✓Because ln 24 is not defined, there is not solution.7. log 3x 5 21 8. 6 ln x 5 30x 5 3 21 ln x 5 5x 5 1 }3x 5 e 5Check: log 3x 5 21 x ø 148.41log 3 1 }1 032 21 Check: 6 ln x 5 309. log 2(x 1 4) 5 5log }1 3}log 3 0 21 6 ln e 5 0 3021 5 21 ✓ 6(5) 0 30x 1 4 5 2 5x 1 4 5 32x 5 28Check: log 2(x 1 4) 5 5log 2(28 1 4) 0 5log 232 0 5log 32}log 2 0 55 5 5 ✓30 5 30 ✓10. (1, 5): 5 5 ab 1 → a 5 5 }b(2, 30): 30 5 ab 230 5 1 }5 b2 b 230 5 5b6 5 ba 5 }5 b 5 }5 6So, an equation is y 5 5 }6 p 6x11. (1, 4): 4 5 ab 1 → a 5 4 }b(2, 32): 32 5 ab 232 5 1 }4 b2 b 232 5 4b8 5 ba 5 }4 b 5 }4 8 5 }1 2So, an equation is y 5 1 }2 p 8x12. (2, 15): 15 5 ab 2 → a 5 15 }b 2(3, 45): 45 5 ab 345 5 1 15 }b 2 2 b345 5 15b3 5 ba 5 }15 b 5 15 2 }3 2 5 }5 3So, an equation is y 5 5 }3 p 3x13. (4, 8): 8 5 a(4) b → a 5 8 }4 b(9, 23): 23 5 a(9) b23 5 1 8 }4 b 2 9b23 5 81 }9 42 b23}8 5 1 }9 42 blog 9/423}8 5 blog }238}log }9 5 b41.30 ø ba 5 }84 ø 8 b } ø 1.321.34So, an equation is y 5 1.32x 1.3Algebra 2Worked-Out Solution Key459