Verified Answer :

  • Establish the domain by ensuring all logarithmic arguments are positive, resulting in \(x > 1.2\).
  • Apply logarithmic power rules to equate the arguments: \(5x + 6 = (5x - 6)^2\).
  • Simplify and solve the resulting quadratic equation: \(5x^2 - 13x + 6 = 0\), yielding \(x = 2\) or \(x = 0.6\).
  • Reject the extraneous solution \(x = 0.6\) as it falls outside the valid domain, confirming the final answer is \(\boxed{2}\).

Explanation

  1. Domain Analysis

To solve the equation \(\log_{10}(5x + 6) = 2 \log_{10}(5x - 6)\), we first identify the domain of the expression. Logarithmic functions \(\log_{10}(u)\) are only defined for \(u > 0\). Thus, we must have \(5x + 6 > 0\) and \(5x - 6 > 0\). Solving these, we find \(x > -1.2\) and \(x > 1.2\). Combining these conditions, the domain of the equation is \(x > 1.2\). Any potential solution must fall within this interval.

  1. Simplifying the Equation

We use the logarithmic power property \(n \log_{b}(u) = \log_{b}(u^n)\) to simplify the equation:
\[\log_{10}(5x + 6) = \log_{10}((5x - 6)^2).\]
Since the logarithm is an injective (one-to-one) function, we can set the arguments equal to each other:
\[5x + 6 = (5x - 6)^2.\]

  1. Forming and Solving the Quadratic Equation

Next, we expand the right side of the equation:
\[5x + 6 = 25x^2 - 60x + 36.\]
Rearranging all terms to one side, we form a quadratic equation:
\[25x^2 - 65x + 30 = 0.\]
To simplify, we divide the entire equation by 5:
\[5x^2 - 13x + 6 = 0.\]

  1. Calculating Potential Solutions

We can solve this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=5\), \(b=-13\), and \(c=6\):
\[x = \frac{13 \pm \sqrt{(-13)^2 - 4(5)(6)}}{2(5)} = \frac{13 \pm \sqrt{169 - 120}}{10} = \frac{13 \pm \sqrt{49}}{10} = \frac{13 \pm 7}{10}.\]
This gives two potential values:
\(x_1 = \frac{13 + 7}{10} = \frac{20}{10} = 2\)
\(x_2 = \frac{13 - 7}{10} = \frac{6}{10} = 0.6\)

  1. Verification and Final Solution

We verify these values against our domain constraint \(x > 1.2\):

  • For \(x = 2\), \(2 > 1.2\) is true.
  • For \(x = 0.6\), \(0.6 > 1.2\) is false, meaning it is an extraneous solution.

Thus, the only valid solution is \(x = 2\). The final answer is \(\boxed{2}\).

Examples

Logarithmic equations like this appear in various scientific fields, including chemistry for calculating pH levels, or in finance for determining the time required for an investment to grow to a certain amount. For instance, if you are modeling the decay of a substance or the growth of a population where rates are compounded continuously, you might encounter an equation that requires you to isolate an unknown variable within a logarithmic function. Solving such equations ensures that you can accurately predict outcomes in systems involving exponential growth or decay.

Related Questions