Question

which equation(s) have x = 2 as the solution?\
$\log_{2}(5x + 6) = 4$\
$\log_{x}16 = 4$\
$\log_{3}(6x + 4) = 3$\
$\log_{x}36 = 6$\
done

Explanation:

Step1: Check \( \log_{2}(5x + 6)=4 \)

Using the definition of logarithms \( \log_{a}b = c \implies b=a^{c} \), we substitute \( x = 2 \):
Left - hand side (LHS): \( 5x+6=5\times2 + 6=10 + 6 = 16 \)
Right - hand side (RHS): \( 2^{4}=16 \)
Since \( LHS = RHS \), \( x = 2 \) is a solution.

Step2: Check \( \log_{x}16 = 4 \)

Using the definition \( \log_{a}b=c\implies b = a^{c} \), we get \( 16=x^{4} \).
Substitute \( x = 2 \): \( x^{4}=2^{4}=16 \), which matches the left - hand side. So \( x = 2 \) is a solution.

Step3: Check \( \log_{3}(6x + 4)=3 \)

Using the definition, \( 6x + 4=3^{3}=27 \)
Substitute \( x = 2 \): \( 6\times2+4=12 + 4 = 16
eq27 \). So \( x = 2 \) is not a solution.

Step4: Check \( \log_{x}36 = 6 \)

Using the definition, \( 36=x^{6} \)
Substitute \( x = 2 \): \( x^{6}=2^{6}=64
eq36 \). So \( x = 2 \) is not a solution.

Answer:

\( \log_{2}(5x + 6)=4 \), \( \log_{x}16 = 4 \)