91Ó°ÊÓ

Logarithms help us solve equations where the unknown appears as an exponent. They have unique properties that make these equations easier to handle.
For example, consider the property: \(\text{log}_b(a) = c\) means \(b^c = a\). This means any logarithm can be rewritten as an exponential equation.
Another important property is the change of base formula: \(\text{log}_b(a) = \frac{\text{log}_k(a)}{\text{log}_k(b)}\), which helps in converting one logarithmic base to another. This can be useful when solving complex logarithmic equations or when using calculators that do not support certain bases.
Understanding these properties is crucial as they form the basis of solving logarithmic equations.

Converting a logarithmic equation to its exponential form is a powerful technique. This is the second step in solving our example exercise.
For the equation \(\text{log}_3(3x - 2) = 2\), using the property \(\text{log}_b(a) = c \implies b^c = a\), we rewrite it as \(3^2 = 3x - 2\).
Here, the base \(b\) is 3, the exponent \(c\) is 2, and \(a\) is the expression inside the logarithm (\(3x - 2\)).
This conversion simplifies the problem by turning it into a form that is easier to solve: a linear equation.

Solving linear equations involves finding the value of the variable that makes the equation true. Let's solve the linear equation from our converted exponential form: \(3^2 = 3x - 2\).
First, calculate the exponent: \(3^2 = 9\). So, we now have \(9 = 3x - 2\).
To isolate \(x\), add 2 to both sides: \(9 + 2 = 3x\). Simplify it to get \(11 = 3x\).
Finally, divide both sides by 3: \(\frac{11}{3} = x\). So, \(x = \frac{11}{3}\).
By isolating the variable step-by-step, we solved the logarithmic equation. This method works for any linear equation: isolate the variable by performing inverse operations.