# 6.4.1 EXERCISESIn Exercises 1 - 24, solve the equation analytically.2. log2 (r*) = log2(r)1. log(3r – 1) = log(4 – r)%3D3. In (8 – a?) = In(2 – 2)4. log, (18 – 2?) = log; (6 – r)6. log (2r – 1) = -35. log3(7 – 2r) = 28. log(r2 – 3r) = 17. In (22 – 99) = 0%3D3x – 2-(")>12. 10 log (0-12)10. log (10-3)9. log125= 4.7%3D2r + 311. – log(r) = 5.4= 15014. 3 In(x) – 2 =1- ln(xr)13. 6 – 3 log; (2x) = 015. log3(r – 4) + log3(r+ 4) = 216. log, (2r + 1) + log, (r+ 2)18. In(r+ 1) – ln(r) = 317. log 169 (3r + 7)- log169(5x-9)%3!%3D19. 2log,(r) = log,(2) + log,(r+ 12)20. log(r) – log(2) = log(r + 8) – log(r + 2)%3D21. log3(x) = log1 (x) + 822. In(In(x)) = 3%3D23. (log(r))? = 2 log(r) + 1524. In(22) = (In(r))²%3D

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I only need help with question 5 help_outlineImage Transcriptionclose6.4.1 EXERCISES In Exercises 1 - 24, solve the equation analytically. 2. log2 (r*) = log2(r) 1. log(3r – 1) = log(4 – r) %3D 3. In (8 – a?) = In(2 – 2) 4. log, (18 – 2?) = log; (6 – r) 6. log (2r – 1) = -3 5. log3(7 – 2r) = 2 8. log(r2 – 3r) = 1 7. In (22 – 99) = 0 %3D 3x – 2 -(")> 12. 10 log (0-12) 10. log (10-3) 9. log125 = 4.7 %3D 2r + 3 11. – log(r) = 5.4 = 150 14. 3 In(x) – 2 =1- ln(xr) 13. 6 – 3 log; (2x) = 0 15. log3(r – 4) + log3(r+ 4) = 2 16. log, (2r + 1) + log, (r+ 2) 18. In(r+ 1) – ln(r) = 3 17. log 169 (3r + 7)- log169(5x-9) %3! %3D 19. 2log,(r) = log,(2) + log,(r+ 12) 20. log(r) – log(2) = log(r + 8) – log(r + 2) %3D 21. log3(x) = log1 (x) + 8 22. In(In(x)) = 3 %3D 23. (log(r))? = 2 log(r) + 15 24. In(22) = (In(r))² %3D fullscreen
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Step 1

Given an equation

Step 2

Now, analytic means to solve the equation using algebraic properties, rules, or theorems. help_outlineImage TranscriptioncloseNote: We change the given logarithmic equation to exponential form. log, (x) = y implies a = x, a > 0, a + 1, y + 0 fullscreen
Step 3

Using the above d... help_outlineImage Transcriptioncloselog, (7 – 2x) = 2 implies 7 – 2x = 32 We now solve the exponential form, 7- 2x = 32 %3D 7- 2x = 9 7-7- 2x = 9 – 7 -2x = 2 -2x -2--2 X = -1 fullscreen

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