Applies all logarithmic properties to solve equations. Covers strategies like condensing, converting to exponential form, using the one-to-one property, and checking for extraneous solutions.
What is the crucial final step after finding a potential solution to a logarithmic equation?
Check for extraneous solutions. You must substitute the solution back into the original equation to ensure that the argument of every logarithm is positive. Any solution that results in taking the log of a negative number or zero must be discarded.
What is the One-to-One Property of Logarithms, and when is it used for solving equations?
If \(\log_b(M) = \log_b(N)\), then \(M = N\). It is used when you have a single logarithm with the same base on both sides of the equation.
Solve for \(x\): \(\log_6(2x - 1) = \log_6(x + 4)\).
Using the One-to-One Property: \(2x - 1 = x + 4\), which gives \(x=5\). Checking the solution: \(\log_6(9)\) and \(\log_6(9)\) are valid. So, \(x=5\) is the solution.
What is the general strategy for solving an equation with multiple logs on one side and a constant on the other, like \(\log_2(x) + \log_2(x-2) = 3\)?
1. Condense the logarithms into a single log. 2. Convert the resulting equation to its exponential form. 3. Solve the resulting equation (often a polynomial). 4. Check for extraneous solutions.
Solve for \(x\): \(\log_2(x) + \log_2(x-2) = 3\).
1. Condense: \(\log_2(x(x-2)) = 3\). 2. Convert: \(x(x-2) = 2^3\). 3. Solve: \(x^2 - 2x = 8 \Rightarrow x^2 - 2x - 8 = 0 \Rightarrow (x-4)(x+2) = 0\). Solutions are \(x=4\) and \(x=-2\). 4. Check: \(x=4\) is valid. \(x=-2\) is extraneous because \(\log_2(-2)\) is undefined.
Solve for \(x\): \(\ln(x) = 5\).
Convert the natural log equation to its exponential form. Remember \(\ln\) is base \(e\). So, \(x = e^5\). Since \(e^5\) is positive, the solution is valid.
Solve for \(x\): \(2\log_5(x) = \log_5(9)\).
Condense the left side using the Power Rule: \(\log_5(x^2) = \log_5(9)\). Use the One-to-One Property: \(x^2 = 9\), so \(x = 3\) or \(x = -3\). Check: \(x=3\) is valid. \(x=-3\) is extraneous because \(\log_5(-3)\) is undefined.
Solve for \(x\): \(\log_4(x+3) = 2\).
Convert to exponential form: \(x+3 = 4^2\). Solve: \(x+3 = 16 \Rightarrow x = 13\). Check: \(\log_4(16)\) is valid. The solution is \(x=13\).
