Logarithm equations I

Technique: converting to exponential form

As we have in an earlier section,

$\log_{a} f (x) = c ⟺ f (x) = a^{c}$

To solve a logarithmic equation, we often simplify our equation so that we only have one logarithm left. We then convert it to exponential form.

Tip: Always check your solutions. The argument of a logarithm must be positive, so we may have to reject answers. For example, if our question has $\ln (x + 2),$ then all answers must be such that $x + 2 > 0.$ That is, $x > - 2.$ Answers like $- 3$ should be rejected.

Example: Solving by converting form

Question

Solve the following equations for $x$ :

$\log_{x} 9 = 2$
$\log_{2} (x - 1) = 3$
$2 \ln x - 1 = 3$
$3 \lg (2 x + 3) + 5 = 2$

Solution

$\begin{matrix} \log_{x} 9 & = 2 \\ x^{2} & = 9 \\ x & = 3 or - 3 (rej. since x > 0) \\ x & = 3 ∎ \end{matrix}$

$\begin{matrix} \log_{2} (x - 1) & = 3 \\ x - 1 & = 2^{3} \\ x - 1 & = 8 \\ x & = 9 ∎ \end{matrix}$

$\begin{matrix} 2 \ln x - 1 & = 3 \\ 2 \ln x & = 4 \\ \ln x & = 2 \\ x & = e^{2} \\ \approx 7.39 ∎ \end{matrix}$

$\begin{matrix} 3 \lg (2 x + 3) + 5 & = 2 \\ 3 \lg (2 x + 3) & = - 3 \\ \lg (2 x + 3) & = - 1 \\ 2 x + 3 & = 10^{- 1} \\ 2 x + 3 & = 0.1 \\ 2 x & = - 2.9 \\ x & = - 1.45 ∎ \end{matrix}$