Question

1. (20 points) solve for the variable.

(a) \\(\ln(x - 2) + \ln(x - 8) = \ln 7\\)

(b) \\(\log_{3}(x) = 3\\)

(c) \\(e^{x} + 40e^{-x} - 13 = 0\\)

(d) \\(8^{-6x} = 7^{-9x - 3}\\)

Explanation:

Step 1: Solve part (a) using logarithm properties.

$$\ln((x-2)(x-8)) = \ln(7) \implies x^2 - 10x + 16 = 7 \implies x^2 - 10x + 9 = 0$$

Step 2: Factor and check domain for part (a).

$$(x-9)(x-1) = 0 \implies x = 9 \text{ or } x = 1 \text{ (extraneous, since } x > 8\text{)} \implies x = 9$$

Step 3: Solve part (b) using exponential form.

$$x = 3^3 \implies x = 27$$

Step 4: Solve part (c) by substituting \(u = e^x\).

$$u + \frac{40}{u} - 13 = 0 \implies u^2 - 13u + 40 = 0$$

Step 5: Factor and solve for \(x\) in part (c).

$$(u-5)(u-8) = 0 \implies e^x = 5 \text{ or } e^x = 8 \implies x = \ln(5) \text{ or } x = \ln(8)$$

Step 6: Solve part (d) by taking natural logarithms.

$$\ln(8^{-6x}) = \ln(7^{-9x-3}) \implies -6x\ln(8) = (-9x-3)\ln(7)$$

Step 7: Isolate \(x\) in part (d).

$$x(9\ln(7) - 6\ln(8)) = -3\ln(7) \implies x = \frac{3\ln(7) + \ln(8)}{9\ln(7) - 6\ln(8)}$$

Answer:

(a) \(x = 9\)
(b) \(x = 27\)
(c) \(x = \ln(5)\) or \(x = \ln(8)\)
(d) \(x = \frac{3\ln(7) + \ln(8)}{9\ln(7) - 6\ln(8)}\)