Solving Logarithmic Equations
Solving logarithmic equations is a Grade 7 math skill from Yoshiwara Intermediate Algebra where students use the definition of logarithms and log properties to isolate the variable. Methods include converting to exponential form and applying the one-to-one property of logarithms.
Key Concepts
Property To solve a logarithmic equation: 1. If the equation contains only one logarithm, convert it to its equivalent exponential form using the definition $y = \log b x \iff x = b^y$. 2. If the equation contains more than one logarithm, use logarithm properties to combine them into a single logarithm first. 3. Always check for extraneous solutions, because the argument of a logarithm must be positive.
Examples To solve $\log 5(x+3) = 2$, convert it to exponential form: $x+3 = 5^2$. This gives $x+3=25$, so $x=22$. The solution is valid.
Solve $\log 3 x + \log 3(x 6) = 3$. First combine the logs: $\log 3(x(x 6)) = 3$. Convert to exponential form: $x(x 6) = 3^3$, which is $x^2 6x=27$. The solutions to $x^2 6x 27=0$ are $x=9$ and $x= 3$. Checking them, $x=9$ is valid, but $x= 3$ is extraneous because $\log 3( 3)$ is undefined.
Common Questions
How do you solve a logarithmic equation?
Isolate the log term, then convert to exponential form: if log_b(x) = c, then x = b^c. Solve for x and check for extraneous solutions.
What is the one-to-one property of logarithms?
If log_b(A) = log_b(B), then A = B. This property allows you to drop the logarithm from both sides when bases match.
How do you solve log(2x + 1) = 3?
Convert to exponential form: 2x + 1 = 10^3 = 1000. Then 2x = 999, so x = 499.5.
Why must you check solutions to log equations?
Logarithms are only defined for positive arguments. A solution that makes the argument zero or negative is extraneous and must be discarded.