Question

you have to wait until you can carve the turkey?
in exercises 21–32, solve the equation. (see example 3.)

1. \\(\ln(2x - 7) = \ln(x + 11)\\)
2. \\(\ln(2x - 4) = \ln(x + 6)\\)
3. \\(\log_2(3x - 4) = \log_2 5\\) 24. \\(\log(7x + 3) = \log 38\\)
4. \\(\log_3(4x + 8) = 5\\) 26. \\(\log_5(2x + 1) = 2\\)
5. \\(\log_4(4x + 9) = 2\\) 28. \\(\log_3(5x + 10) = 4\\)
6. \\(\log(12x - 9) = \log 3x\\) 30. \\(\log_4(5x + 9) = \log_4 6x\\)
7. \\(\log_2(x^2 - x - 6) = 2\\) 32. \\(\log_3(x^2 + 9x + 27) = 2\\)

Explanation:

Problem 21: $\ln(4x - 7) = \ln(x + 11)$

Step1: Set arguments equal

Since $\ln a = \ln b$ implies $a=b$ (for $a,b>0$):
$4x - 7 = x + 11$

Step2: Isolate $x$

Subtract $x$ and add 7 to both sides:
$4x - x = 11 + 7$
$3x = 18$
$x = 6$

Step3: Verify domain

$4(6)-7=17>0$, $6+11=17>0$, valid.

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Problem 22: $\ln(2x - 4) = \ln(x + 6)$

Step1: Set arguments equal

$2x - 4 = x + 6$

Step2: Isolate $x$

$2x - x = 6 + 4$
$x = 10$

Step3: Verify domain

$2(10)-4=16>0$, $10+6=16>0$, valid.

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Problem 23: $\log_2(3x - 4) = \log_2 5$

Step1: Set arguments equal

$3x - 4 = 5$

Step2: Isolate $x$

$3x = 5 + 4$
$3x = 9$
$x = 3$

Step3: Verify domain

$3(3)-4=5>0$, valid.

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Problem 24: $\log(7x + 3) = \log 38$

Step1: Set arguments equal

$7x + 3 = 38$

Step2: Isolate $x$

$7x = 38 - 3$
$7x = 35$
$x = 5$

Step3: Verify domain

$7(5)+3=38>0$, valid.

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Problem 25: $\log_2(4x + 8) = 5$

Step1: Rewrite in exponential form

$\log_b a = c \implies b^c = a$:
$2^5 = 4x + 8$
$32 = 4x + 8$

Step2: Isolate $x$

$4x = 32 - 8$
$4x = 24$
$x = 6$

Step3: Verify domain

$4(6)+8=32>0$, valid.

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Problem 26: $\log_3(2x + 1) = 2$

Step1: Rewrite in exponential form

$3^2 = 2x + 1$
$9 = 2x + 1$

Step2: Isolate $x$

$2x = 9 - 1$
$2x = 8$
$x = 4$

Step3: Verify domain

$2(4)+1=9>0$, valid.

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Problem 27: $\log(4x + 9) = 2$

Step1: Rewrite in exponential form (base 10)

$10^2 = 4x + 9$
$100 = 4x + 9$

Step2: Isolate $x$

$4x = 100 - 9$
$4x = 91$
$x = \frac{91}{4} = 22.75$

Step3: Verify domain

$4(22.75)+9=100>0$, valid.

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Problem 28: $\log_4(5x + 1) = 2$

Step1: Rewrite in exponential form

$4^2 = 5x + 1$
$16 = 5x + 1$

Step2: Isolate $x$

$5x = 16 - 1$
$5x = 15$
$x = 3$

Step3: Verify domain

$5(3)+1=16>0$, valid.

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Problem 29: $\log(12x - 9) = \log 3x$

Step1: Set arguments equal

$12x - 9 = 3x$

Step2: Isolate $x$

$12x - 3x = 9$
$9x = 9$
$x = 1$

Step3: Verify domain

$12(1)-9=3>0$, $3(1)=3>0$, valid.

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Problem 30: $\log_3(5x + 9) = \log_3 6x$

Step1: Set arguments equal

$5x + 9 = 6x$

Step2: Isolate $x$

$6x - 5x = 9$
$x = 9$

Step3: Verify domain

$5(9)+9=54>0$, $6(9)=54>0$, valid.

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Problem 31: $\log_2(x^2 - x - 6) = 2$

Step1: Rewrite in exponential form

$2^2 = x^2 - x - 6$
$4 = x^2 - x - 6$

Step2: Rearrange to quadratic

$x^2 - x - 6 - 4 = 0$
$x^2 - x - 10 = 0$

Step3: Solve quadratic

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=1,b=-1,c=-10$:
$x = \frac{1\pm\sqrt{1 + 40}}{2} = \frac{1\pm\sqrt{41}}{2}$

Step4: Verify domain

$x^2 - x -6>0$:

• $\frac{1+\sqrt{41}}{2}\approx3.701$: $(3.701)^2 - 3.701 -6\approx13.7-3.7-6=4>0$ (valid)
• $\frac{1-\sqrt{41}}{2}\approx-2.701$: $(-2.701)^2 - (-2.701) -6\approx7.3+2.7-6=4>0$ (valid)

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Problem 32: $\log_5(x^2 + 9x + 27) = 2$

Step1: Rewrite in exponential form

$5^2 = x^2 + 9x + 27$
$25 = x^2 + 9x + 27$

Step2: Rearrange to quadratic

$x^2 + 9x + 27 - 25 = 0$
$x^2 + 9x + 2 = 0$

Step3: Solve quadratic

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=1,b=9,c=2$:
$x = \frac{-9\pm\sqrt{81 - 8}}{2} = \frac{-9\pm\sqrt{73}}{2}$

Step4: Verify domain

$x^2 +9x +27>0$: discriminant of quadratic is $81-108=-27<0$, so it is always positive (valid for both roots).

Answer:

1. $x=6$
2. $x=10$
3. $x=3$
4. $x=5$
5. $x=6$
6. $x=4$
7. $x=\frac{91}{4}$
8. $x=3$
9. $x=1$
10. $x=9$
11. $x=\frac{1+\sqrt{41}}{2}, x=\frac{1-\sqrt{41}}{2}$
12. $x=\frac{-9+\sqrt{73}}{2}, x=\frac{-9-\sqrt{73}}{2}$