Q. Select all the solutions of the equation \(2\log x = \log(5x – 4)\).

Q: How do I solve 2\log x = \log(5x-4)?

Use 2\log x=\log x^2, so \log x^2=\log(5x-4)\Rightarrow x^2=5x-4\Rightarrow x^2-5x+4=0\Rightarrow (x-1)(x-4)=0, giving x=1 or x=4. Check domain after.

Q: What domain restrictions apply?

Arguments must be positive: x>0 and 5x-4>0, so x>\frac{4}{5}. Only roots greater than 4/5 are valid.

Q: Why can't x=0 or x=-1 be solutions?

\log x is undefined for x \le 0 , so x=0 and x=-1 are not in the domain and cannot satisfy the equation.

Q: Is it valid to rewrite 2\log x as \log x^2?.

Yes, for x>0 the log power rule gives 2\log x = \log(x^2) for any log base, so the step is valid under the domain constraint.

Q: How can I verify the solutions by substitution?

For x=1: 2\log 1=0 and \log(5-4)=0. For x=4: 2\log 4=\log 16 and \log(20-4)=\log 16. Both check out.

Answer

\(2\log x=\log(5x-4)\Rightarrow \log(x^2)=\log(5x-4)\), so \(x^2=5x-4\). Domain: \(x>0\) and \(5x-4>0\Rightarrow x>4/5\). Solve \(x^2-5x+4=0\): \(x=1\) or \(x=4\), both valid.

Answer: a and b.

Detailed Explanation

Problem

Solve the equation \(2\log x = \log(5x – 4)\) and select all solutions from the list: a. \(x=1\), b. \(x=4\), c. \(x=5\), d. \(x=0\), e. \(x=-1\).

Domain (where both logarithms are defined):For \(\log x\) we need \(x>0\). For \(\log(5x-4)\) we need \(5x-4>0\), which is \(x>\tfrac{4}{5}\). Both conditions together give the domain \(x>\tfrac{4}{5}\).
Use logarithm property to combine the left side:The property \(n\log a = \log(a^n)\) gives \(2\log x = \log(x^2)\). So the equation becomes
\(\log(x^2) = \log(5x-4)\).
Equate arguments (same base logs):Since the logs (same base) are equal and the expressions are in the domain, the arguments are equal:
\(x^2 = 5x – 4\).
Solve the resulting quadratic:Rearrange:
\(x^2 – 5x + 4 = 0\).

Factor:

\((x-1)(x-4) = 0\).

Thus the algebraic solutions are \(x=1\) and \(x=4\).
Check against the domain:Both \(x=1\) and \(x=4\) satisfy \(x>\tfrac{4}{5}\), so both are valid solutions of the original equation.
Conclusion (which choices are correct):The solutions are \(x=1\) and \(x=4\). Therefore the correct choices are a and b.

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Algebra FAQs

How do I solve \(2\log x = \log(5x-4)\)?

Use \(2\log x=\log x^2\), so \(\log x^2=\log(5x-4)\Rightarrow x^2=5x-4\Rightarrow x^2-5x+4=0\Rightarrow (x-1)(x-4)=0\), giving \(x=1\) or \(x=4\). Check domain after.

What domain restrictions apply?

Arguments must be positive: \(x>0\) and \(5x-4>0\), so \(x>\frac{4}{5}\). Only roots greater than \(4/5\) are valid.

Which answer choices are correct?

a. \(x = 1\) and b. \(x = 4\).

Why can't \(x=0\) or \(x=-1\) be solutions?

\( \log x \) is undefined for \( x \le 0 \), so \( x=0 \) and \( x=-1 \) are not in the domain and cannot satisfy the equation.

Is it valid to rewrite \(2\log x\) as \(\log x^2\)?.

Yes, for \(x>0\) the log power rule gives \(2\log x = \log(x^2)\) for any log base, so the step is valid under the domain constraint.

Could the logarithm base affect the solution?

Do we need to check for extraneous roots?

Always. Solving algebraically can produce roots violating domain restrictions; here both \(x=1\) and \(x=4\) satisfy \(x>4/5\), so neither is extraneous.

How can I verify the solutions by substitution?

For \(x=1\): \(2\log 1=0\) and \(\log(5-4)=0\). For \(x=4\): \(2\log 4=\log 16\) and \(\log(20-4)=\log 16\). Both check out.

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