By Chris Deziel, Updated Aug 30, 2022
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Logarithms can turn an otherwise straightforward algebraic problem into a knotty one. They are often viewed as tedious, hard to manipulate, and somewhat mysterious. The good news is that removing them from an equation is straightforward once you remember that a logarithm is simply the inverse of an exponent.
While the base of a logarithm can be any positive number, the most common bases in science are 10 and Euler’s number e. In mathematics, “log” denotes a base‑10 logarithm and “ln” denotes a natural logarithm with base e.
To eliminate logarithms, raise both sides of the equation to the same power as the logarithm’s base. If the equation contains several logarithms, move all of them to one side and simplify first.
A logarithm answers the question “to what power must the base be raised to produce a given number?” In other words, the logarithm of a number is the exponent required to obtain that number from the base. For example, \(\log_8 2 = 6\) means that 82 = 64. In the common notation \(\log x = 100\), the base is understood to be 10, so the question becomes “10 raised to what power equals 100?” The answer is 2, because 102 = 100.
Because a logarithm is the inverse operation of exponentiation, equations containing logarithms can often be “untangled” by applying the appropriate exponent to both sides. This works as long as all logarithms involved share the same base.
Simple logarithm
\(\log x = y\)
Raise both sides to the power of 10: \(10^{\log x} = 10^y\). Since 10^{\log x} = x, we get \(x = 10^y\).
All terms are logarithms
\(\log (x^2 - 1) = \log (x + 1)\)
Exponentiate both sides with base 10: \(x^2 - 1 = x + 1\). Simplify to obtain \(x^2 - x - 2 = 0\), whose solutions are \(x = -2\) or \(x = 1\).
Mixed logarithms and algebraic terms
Follow these steps:
1. Start with the equation, for example: \(\log x = \log (x - 2) + 3\).
2. Move all logarithms to one side: \(\log x - \log (x - 2) = 3\).
3. Apply the logarithm laws: \(\log \left(\frac{x}{x-2}\right) = 3\).
4. Exponentiate both sides with base 10: \(\frac{x}{x-2} = 10^3\).
5. Solve for x: \(x = 1000x - 2000 \Rightarrow -999x = -2000 \Rightarrow x = \frac{2000}{999} \approx 2.002\).
By systematically applying these rules, you can eliminate logarithms from almost any algebraic equation.
