Logarithmic equations

To solve an equation in which the unknown appears in the argument of a logarithm one follows a fixed scheme, which always starts from the existence conditions.

In summary — Methods

1. Write the ECs (arguments $>0$) first of all.
2. Reduce to a single logarithm per side using the properties, then exploit injectivity: $\log_a f(x) = \log_a g(x) \iff f(x) = g(x)$ (with ECs).
3. Substitution: if $\log_a x$ appears several times, set $t = \log_a x$ and solve in $t$.
4. Passing to the exponential: $\log_a f(x) = k \iff f(x) = a^k$.
5. Check the ECs always at the end.

Example — $\log_3(x+1) + \log_3(x-1) = 1$

EC: $x+1 > 0 \ \wedge\ x-1 > 0 \iff x > 1$.

Union of the logarithms (product property): $\log_3\bigl((x+1)(x-1)\bigr) = 1$, that is $\log_3(x^2-1) = 1$.

Passing to the exponential: $x^2 - 1 = 3^1 = 3 \iff x^2 = 4 \iff x = \pm 2$.

EC check: $x = -2$ does not satisfy $x > 1$ (discarded); $x = 2$ does.

Solution: $\boxed{x = 2}$.

Links

Topics: Logarithmic function
Concepts: Existence conditions · Logarithmic equations · Injectivity · Logarithm
Functions: Logarithmic function
Methods: Logarithm equations
Skills: Solve equations