Focuses on strategies for solving logarithmic equations. This includes converting to exponential form, using log properties to condense expressions, and the critical step of checking for extraneous solutions.
What is the first step to solve a logarithmic equation with multiple logarithms on one side, like \(\log_2(x) + \log_2(x-2) = 3\)?
Use the properties of logarithms (in this case, the Product Rule) to condense the logarithms into a single logarithmic expression. The equation becomes \(\log_2(x(x-2)) = 3\).
After condensing a logarithmic equation into the form \(\log_b(A) = C\), what is the next step?
Rewrite the equation in its equivalent exponential form, \(b^C = A\), and then solve for the variable.
Solve for \(x\): \(\log_4(x+3) = 2\).
Rewrite in exponential form: \(4^2 = x+3\). This gives \(16 = x+3\). Solving for \(x\) yields \(x=13\). Always check that the argument \((x+3)\) is positive. For \(x=13\), \(13+3=16 > 0\), so the solution is valid.
What property can be used to solve an equation of the form \(\log_b(M) = \log_b(N)\)?
The one-to-one property of logarithms. If the bases are the same, you can set the arguments equal to each other: \(M=N\).
Solve for \(x\): \(\ln(3x - 4) = \ln(x + 6)\).
Using the one-to-one property, set the arguments equal: \(3x - 4 = x + 6\). This gives \(2x = 10\), so \(x=5\). Checking the arguments: \(3(5)-4 = 11 > 0\) and \(5+6 = 11 > 0\). The solution is valid.
Solve for \(x\): \(\log(x) + \log(x+21) = 2\).
1. Condense: \(\log(x(x+21)) = 2\).
2. Convert to exponential form (base is 10): \(10^2 = x(x+21)\).
3. Solve the quadratic: \(100 = x^2+21x \implies x^2+21x-100 = 0 \implies (x+25)(x-4) = 0\). Solutions are \(x=-25\) and \(x=4\).
4. Check for extraneous solutions: \(x=-25\) is extraneous. \(x=4\) is valid.
Why must all potential solutions to a logarithmic equation be checked?
To check for extraneous solutions. The domain of a logarithm \(\log_b(A)\) is \(A>0\). Algebraic steps can produce solutions that violate this domain restriction. Any solution that results in taking the log of a non-positive number must be discarded.
What is an extraneous solution in the context of logarithmic equations?
An extraneous solution is a result that is correctly derived from algebraic manipulation of the equation, but does not satisfy the original equation because it falls outside the domain of one of the original logarithmic expressions.
Solve for \(x\): \(\log_3(2x-1) - \log_3(x-4) = 2\).
1. Condense: \(\log_3\left(\frac{2x-1}{x-4}\right) = 2\).
2. Convert to exponential form: \(3^2 = \frac{2x-1}{x-4}\).
3. Solve: \(9 = \frac{2x-1}{x-4} \implies 9(x-4) = 2x-1 \implies 9x-36 = 2x-1 \implies 7x=35 \implies x=5\).
4. Check: \(2(5)-1=9>0\) and \(5-4=1>0\). The solution is valid.
