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Logarithms — Solving logarithmic equations 7 problems

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📘 Notes

Logarithms — Solving Logarithmic Equations

Logarithmic equations can be solved by using logarithm laws or by changing the equation into exponential form.

1. Key logarithm facts

For logarithms, remember:

\[ \log a+\log b=\log(ab) \] \[ \log a-\log b=\log\left(\frac{a}{b}\right) \] \[ n\log a=\log(a^n) \]

Also:

\[ \log_a x = y \quad \Longleftrightarrow \quad a^y=x \]

2. Important restriction

A logarithm is only defined when its argument is positive.

\[ \log x \text{ exists only if } x>0 \]

Always check that your final answers make every logarithm valid.

3. Method 1: Use log laws first

Example: Solve

\[ \log x+\log (x-3)=1 \]

Combine the logarithms:

\[ \log\bigl(x(x-3)\bigr)=1 \]

If the logarithm is base 10, then:

\[ x(x-3)=10^1=10 \]

\[ x^2-3x-10=0 \] \[ (x-5)(x+2)=0 \]

So \(x=5\) or \(x=-2\).

Check the restriction:

  • \(x=5\): valid
  • \(x=-2\): not valid because \(\log x\) does not exist
\[ x=5 \]

4. Method 2: Equate arguments

Example: Solve

\[ \log(2x+1)=\log 7 \]

If the bases are the same, then the arguments are equal:

\[ 2x+1=7 \] \[ 2x=6 \] \[ x=3 \]

\[ x=3 \]

5. Method 3: Change to exponential form

Example: Solve

\[ \log_2(x+1)=3 \]

Rewrite in exponential form:

\[ x+1=2^3 \] \[ x+1=8 \] \[ x=7 \]

\[ x=7 \]

6. Example with subtraction

Example: Solve

\[ \log(x+4)-\log x=\log 2 \]

Use the subtraction law:

\[ \log\left(\frac{x+4}{x}\right)=\log 2 \]

So:

\[ \frac{x+4}{x}=2 \] \[ x+4=2x \] \[ x=4 \]

Check: \(x>0\), so the answer is valid.

\[ x=4 \]

7. Exam tips

  • Always check that everything inside a logarithm is positive.
  • Use log laws to combine logarithms before solving.
  • If \(\log a=\log b\), then \(a=b\).
  • If needed, rewrite in exponential form.
  • Check your final answers, because logarithmic equations can produce invalid roots.
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