Answer :

To solve the equation [tex]2 \log_5(3x-2) - \log_5 x = 2[/tex] using the laws of logarithms, follow these steps:


  1. Apply the Power Rule of Logarithms: The power rule states that [tex]a \log_b c = \log_b c^a[/tex]. Therefore, [tex]2 \log_5(3x-2)[/tex] can be rewritten as:

    [tex]\log_5((3x-2)^2)[/tex]


  2. Combine the Logarithms Using the Quotient Rule: The quotient rule states that [tex]\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)[/tex]. Apply this to your equation:

    [tex]\log_5((3x-2)^2) - \log_5 x = \log_5\left(\frac{(3x-2)^2}{x}\right)[/tex]

    So the equation becomes:

    [tex]\log_5\left(\frac{(3x-2)^2}{x}\right) = 2[/tex]


  3. Convert the Logarithmic Equation to an Exponential Equation: The equation [tex]\log_b A = C[/tex] translates to [tex]A = b^C[/tex]. Therefore, the equation [tex]\log_5\left(\frac{(3x-2)^2}{x}\right) = 2[/tex] becomes:

    [tex]\frac{(3x-2)^2}{x} = 5^2[/tex]

    Simplify the right side:

    [tex]\frac{(3x-2)^2}{x} = 25[/tex]


  4. Solve for [tex]x[/tex]:

    Multiply both sides by [tex]x[/tex] to clear the fraction:

    [tex](3x-2)^2 = 25x[/tex]

    Expand the left side:\

    [tex]9x^2 - 12x + 4 = 25x[/tex]

    Rearrange the equation to set it to zero:

    [tex]9x^2 - 12x + 4 - 25x = 0[/tex]

    [tex]9x^2 - 37x + 4 = 0[/tex]


  5. Use the Quadratic Formula: The quadratic formula [tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex] can be used here where [tex]a = 9[/tex], [tex]b = -37[/tex], and [tex]c = 4[/tex].

    Substitute these values into the quadratic formula:

    [tex]x = \frac{-(-37) \pm \sqrt{(-37)^2 - 4 \cdot 9 \cdot 4}}{2 \cdot 9}[/tex]

    Simplify:

    [tex]x = \frac{37 \pm \sqrt{1369 - 144}}{18}[/tex]

    [tex]x = \frac{37 \pm \sqrt{1225}}{18}[/tex]

    [tex]x = \frac{37 \pm 35}{18}[/tex]

    Therefore, the solutions are:

    [tex]x = \frac{37 + 35}{18} = 4[/tex]
    [tex]x = \frac{37 - 35}{18} = \frac{2}{18} = \frac{1}{9}[/tex]


  6. Check for Valid Solutions: These [tex]x[/tex] values must make the original logarithmic expressions valid (i.e., their arguments can't be non-positive).


    • For [tex]x = 4[/tex]: Substitute into [tex]3x - 2[/tex] gives [tex]3(4) - 2 = 10[/tex]. Both [tex]3x-2[/tex] and [tex]x[/tex] are positive, so [tex]x = 4[/tex] is valid.

    • For [tex]x = \frac{1}{9}[/tex]: Substitute into [tex]3x - 2[/tex] gives [tex]3\left(\frac{1}{9}\right) - 2 = \frac{1}{3} - 2[/tex], which is negative, so [tex]x = \frac{1}{9}[/tex] is not valid.




Thus, the only solution is [tex]x = 4[/tex].

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