Q. What is the solution to \(2\log_{5}x=\log_{5}4\)? (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
\[
x=2
\]
See full solution
FAQs
How do you solve 2log_5 x = log_5 4?
Combine: 2log_5 x = log_5 x^2 = log_5 4, so x^2 = 4. Solve x = ±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 nlog_a b = log_a(b^n) and if log_a A = log_a B (a>0, a≠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 = tfrac{1}{2}log_5 4 and x = 5^y = 5^{tfrac{1}{2}log_5 4} = 4^{1/2} = 2.
What if the bases differ on each side?
Use change-of-base: log_b A = frac{log_c A}{log_c b} (e.g., natural logs). Convert both sides to the same base, then combine and solve.
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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