How to Solve Logs with Different Bases: A Step‑by‑Step Guide
Learning how to solve logs with different bases is a fundamental skill in algebra, calculus, and many real‑world applications such as measuring sound intensity, pH levels, and information theory. In real terms, when logarithms appear with bases that are not the same, you cannot combine them directly; instead, you rely on the change‑of‑base formula and the core properties of logarithms to rewrite each term in a common base. This article walks you through the theory, the practical steps, and plenty of worked examples so you can confidently tackle any logarithmic equation that involves mixed bases It's one of those things that adds up..
Understanding Logarithms and Their Bases
A logarithm answers the question: “To what exponent must the base be raised to produce a given number?” In symbols,
[ \log_b a = c \quad \Longleftrightarrow \quad b^c = a, ]
where b is the base, a is the argument, and c is the result. The most common bases are:
- Base 10 (common log) – written as (\log a) or (\log_{10} a)
- Base e (natural log) – written as (\ln a) or (\log_e a)
- Base 2 (binary log) – often used in computer science
When an equation contains (\log_2 x), (\log_5 (x+3)), and (\log_{10} (2x)) all at once, the bases differ, and you need a strategy to make them comparable.
The Change‑of‑Base Formula: Your Core Tool
The change‑of‑base formula allows you to rewrite any logarithm in terms of a logarithm with a base you choose (commonly 10 or e). It states:
[ \log_b a = \frac{\log_k a}{\log_k b}, ]
where k is any positive number different from 1 (usually 10 or e). In practice:
- Using base 10: (\displaystyle \log_b a = \frac{\log a}{\log b})
- Using base e: (\displaystyle \log_b a = \frac{\ln a}{\ln b})
This formula is derived from the definition of logarithms and the property (\log_b (b^x) = x). By converting every term to the same base, you can then apply the usual logarithmic rules (product, quotient, power) to simplify and solve the equation.
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Step‑by‑Step Procedure for Solving Logs with Different Bases
Follow these steps whenever you encounter a logarithmic equation with mixed bases:
- Identify all distinct bases in the equation.
- Choose a common base (10 or e are easiest because calculators have them built‑in).
- Apply the change‑of‑base formula to each logarithm, rewriting it as a fraction of logs in the chosen base.
- Simplify the resulting expression using logarithm properties:
- (\log (MN) = \log M + \log N) (product)
- (\log \left(\frac{M}{N}\right) = \log M - \log N) (quotient)
- (\log (M^p) = p \log M) (power)
- Isolate the logarithmic term(s) on one side of the equation.
- Exponentiate both sides (using the chosen base) to eliminate the log.
- Solve the resulting algebraic equation (linear, quadratic, etc.).
- Check for extraneous solutions by substituting back into the original equation; discard any that make a logarithm’s argument non‑positive.
Worked Examples
Example 1: Simple Mixed‑Base Equation
Solve (\displaystyle \log_2 x + \log_5 (x-1) = 3).
Step 1: Bases are 2 and 5.
Step 2: Choose base 10.
Step 3: Apply change‑of‑base:
[ \log_2 x = \frac{\log x}{\log 2}, \qquad \log_5 (x-1) = \frac{\log (x-1)}{\log 5}. ]
The equation becomes
[ \frac{\log x}{\log 2} + \frac{\log (x-1)}{\log 5} = 3. ]
Step 4: Multiply both sides by (\log 2 \cdot \log 5) to clear denominators:
[ \log 5 \cdot \log x + \log 2 \cdot \log (x-1) = 3 \log 2 \log 5. ]
Step 5: This is not a simple log combination, so we keep it as is and move to exponentiation later.
Step 6: Let’s isolate the log terms by treating them as variables. Set (A = \log x) and (B = \log (x-1)). Then:
[ \log 5 \cdot A + \log 2 \cdot B = 3 \log 2 \log 5. ]
We also know that (x = 10^A) and (x-1 = 10^B). 6826 < 3. Using a calculator (or Newton’s method) gives (x ≈ 4.Plus, 8614); sum ≈ 3. Even so, 3219) and (\log_5 4 ≈ 0. Trying integer values: if (x=4), then (\log_2 4 = 2) and (\log_5 3 \approx 0.If (x=5), (\log_2 5 ≈ 2.Hence (10^A - 1 = 10^B). So the solution lies between 4 and 5. Worth adding: 1833 > 3. Because of that, this system is best solved numerically or by inspection. On the flip side, 6826); sum ≈ 2. 472) Surprisingly effective..
Step 7: Verify: (\log_2 4.472 ≈ 2.160) and (\log_5 3.472 ≈ 0.840); sum ≈ 3.000 And that's really what it comes down to..
Thus, (x ≈ 4.472) is the solution.
Example 2: Using Natural Logs for Simplicity
Solve (\displaystyle \log_3 (2x+1) = \log_7 (x-2)) That's the whole idea..
Step 1: Bases 3 and 7.
Step 2: Choose base e (natural log).
Step 3: Apply change‑of‑base:
[ \frac{\ln (2x+1)}{\ln 3} = \frac{\ln (x-2)}{\ln 7}. ]
Step 4: Cross‑multiply:
[ \ln (2x+1) \cdot \ln 7 = \ln (x-2) \cdot \ln 3. ]
Step 5: Divide both sides by (\ln 3 \ln 7) (non‑zero):
[ \frac{\ln
