Q. What is the solution to \(2\log_5(x)=\log_5(4)\)?

Q: How do you solve 2log_5 x = log_5 4?

Combine: 2\log_5(x) = \log_5(x^2) = \log_5(4), so x^2 = 4. Solve x = \pm 2, but domain requires x>0, so x = 2.

Q: What log rules were used?

Used n\log_a(b) = \log_a(b^n) and if \log_a(A) = \log_a(B) (a>0, a \neq 1) then A = B.

Q: Could you solve it by substituting a variable?

Yes. Let y = \log_5(x), then 2y = \log_5(4) so y = \frac{1}{2}\log_5(4) and x = 5^y = 5^{\frac{1}{2}\log_5(4)} = 4^{1/2} = 2.

Q: What is the domain of the original equation?

Domain: x > 0, since \log_5(x) is defined only for positive x.

Answer

Start with the equation:
\(2\log_5 x = \log_5 4\)
Divide both sides by 2:
\(\log_5 x = \tfrac{1}{2}\log_5 4\)
Use the power rule for logarithms \(\alpha\log_b c=\log_b c^{\alpha}\):
\(\log_5 x = \log_5 4^{1/2} = \log_5 2\)
Since \(\log_5\) is one-to-one, the arguments are equal:
\(x = 2\)
Final result: \(\boxed{2}\)

Detailed Explanation

Problem: Solve the equation \( 2 \log_{5} x = \log_{5} 4 \)

Step 1 – Apply the logarithm power rule

Bring the constant factor 2 inside the logarithm as an exponent:

\[
2\log_{5}x=\log_{5}\bigl(x^{2}\bigr)
\]

Step 2 – Substitute into the equation

\[
\log_{5}\bigl(x^{2}\bigr)=\log_{5}4
\]

Step 3 – Use the one-to-one property of logarithms

Equal logarithms with the same base imply equal arguments:

\[
x^{2}=4
\]

Step 4 – Solve the quadratic

\[
x^{2}=4\quad\Rightarrow\quad x=2\text{ or }x=-2
\]

Step 5 – Check the domain

Logarithms require positive arguments, so \(x>0\). Therefore \(x=-2\) is not allowed.

Answer. So, what is the solution to 2log5x = log54? x =

\[
x=2
\]

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FAQs

How do you solve 2log_5 x = log_5 4?

Combine: \(2\log_5(x) = \log_5(x^2) = \log_5(4)\), so \(x^2 = 4\). Solve \(x = \pm 2\), but domain requires \(x>0\), so \(x = 2\).

Why is x = -2 not allowed?

log_5(-2) is undefined because logarithms require positive arguments. Squaring introduced an extraneous negative root; only positive solutions count.

What log rules were used?

Used \(n\log_a(b) = \log_a(b^n)\) and if \(\log_a(A) = \log_a(B)\) (\(a>0, a \neq 1\)) then \(A = B\).

How can you check the solution?

Substitute x = 2: 2log_5 2 = 2cdotlog_5 2 and log_5 4 = log_5(2^2) = 2log_5 2, so both sides match.

Could you solve it by substituting a variable?

Yes. Let \(y = \log_5(x)\), then \(2y = \log_5(4)\) so \(y = \frac{1}{2}\log_5(4)\) and \(x = 5^y = 5^{\frac{1}{2}\log_5(4)} = 4^{1/2} = 2\).

What if the bases differ on each side?

What is the domain of the original equation?

Domain: \(x > 0\), since \(\log_5(x)\) is defined only for positive \(x\).

General tips for solving log equations?

Combine logs using log rules, convert to a single log or exponential form, solve algebraically, and always check domain to reject extraneous roots.

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