Question

select all of the potential solutions of the equation 2log₅x = log₅4. x=-2 x=4 x=36 x=-36 x=2

Explanation:

Step1: Use Logarithm Property

Recall the logarithm power rule: \( n\log_b a=\log_b a^n \). Apply it to the left - hand side of the equation \( 2\log_2 x=\log_2 4 \), we get \( \log_2 x^2=\log_2 4 \).

Step2: Set Arguments Equal

Since the logarithmic function \( y = \log_b u \) is one - to - one (if \( \log_b u=\log_b v \), then \( u = v \) for \( u>0,v>0,b>0,b
eq1 \)), we can set the arguments equal: \( x^2 = 4 \).

Step3: Solve the Quadratic Equation

Solving \( x^2=4 \), we have \( x=\pm2 \). But we must consider the domain of the logarithmic function \( y = \log_2 x \). The argument of a logarithm must be positive, so \( x>0 \).

Step4: Check the Solutions

For \( x = 2 \): \( \log_2 2=1 \), so \( 2\log_2 2 = 2\times1=2 \), and \( \log_2 4=\log_2 2^2 = 2 \). So \( x = 2 \) is a solution.
For \( x=- 2 \): \( \log_2(-2) \) is undefined (since the argument of a logarithm must be positive), so \( x=-2 \) is not a solution.
For \( x = 4 \): \( 2\log_2 4=2\times2 = 4 \), and \( \log_2 4 = 2 \), \( 4
eq2 \), so \( x = 4 \) is not a solution.
For \( x = 36 \): \( 2\log_2 36\approx2\times5.17 = 10.34 \), and \( \log_2 4 = 2 \), \( 10.34
eq2 \), so \( x = 36 \) is not a solution.
For \( x=-36 \): \( \log_2(-36) \) is undefined, so \( x=-36 \) is not a solution.

Answer:

\( x = 2 \)