Answer:
[tex]x \approx -0.208[/tex]
Step-by-step explanation:
By applying logarithms to each side of the equation and some algebraic handling:
[tex]x+2 = \log_{4} 12[/tex]
[tex]x + 2 = \frac{\log_{10} 12}{\log_{10} 4}[/tex]
[tex]x = \frac{\log_{10}12}{\log_{10}4} - 2[/tex]
[tex]x \approx -0.208[/tex]
