Question

in exercises 33–40, solve the equation. check for extraneous solutions. (see example 4.)

1. \\(\log_6 x + \log_6 (x - 2) = 3\\)
2. \\(\log_5 3x + \log_5 (x - 1) = 3\\)
3. \\(\ln x + \ln (x + 3) = 4\\)
4. \\(\ln x + \ln (x - 2) = 5\\)
5. \\(\log_4 3x^2 + \log_4 3 = 2\\)
6. \\(\log_4 (-x) + \log_4 (x + 10) = 2\\)

Explanation:

Problem 33: $\log_{2}x + \log_{2}(x-2) = 3$

Step1: Combine logs via product rule

$\log_{2}[x(x-2)] = 3$

Step2: Convert to exponential form

$x(x-2) = 2^3 = 8$

Step3: Expand and rearrange to quadratic

$x^2 - 2x - 8 = 0$

Step4: Factor quadratic equation

$(x-4)(x+2) = 0$

Step5: Solve for x, check domain

$x=4$ (valid, $x>2$); $x=-2$ (invalid, log of negative)

Step1: Combine logs via product rule

$\log_{3}[x(x-1)] = 3$

Step2: Convert to exponential form

$x(x-1) = 3^3 = 27$

Step3: Expand and rearrange to quadratic

$x^2 - x - 27 = 0$

Step4: Apply quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x = \frac{1\pm\sqrt{1 + 108}}{2} = \frac{1\pm\sqrt{109}}{2}$

Step5: Check domain (x>1)

Only $x=\frac{1+\sqrt{109}}{2}$ is valid

Step1: Combine logs via product rule

$\ln[x(x+3)] = 4$

Step2: Convert to exponential form

$x(x+3) = e^4$

Step3: Expand and rearrange to quadratic

$x^2 + 3x - e^4 = 0$

Step4: Apply quadratic formula

$x = \frac{-3\pm\sqrt{9 + 4e^4}}{2}$

Step5: Check domain (x>0)

Only $x=\frac{-3+\sqrt{9 + 4e^4}}{2}$ is valid

Answer:

$x=4$

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Problem 34: $\log_{3}x + \log_{3}(x-1) = 3$