Question

Solve the equation. Check for extraneous solutions
log base of 4 (x+3) =-2
Help!

Answer 1

\[ \begin{split}
log_{4}(x+3) &= -2\\
4^{log_{4}(x+3)} &= 4^{-2}\\
x+3 &= \frac{1}{4^2}\\
x &= \frac{1}{16} - 3\\
x &= \frac{1}{16} - \frac{48}{16}\\
x &= \frac{1 - 48}{16}\\
x &= \frac{-47}{16}\\
\end{split} \]
Check:
\[ \begin{split}
log_{4}(\frac{-47}{16} + 3) = -2\\
log_{4}(\frac{-47}{16} + \frac{48}{16}) = -2\\
log_{4}(\frac{-47 + 48}{16}) = -2\\
log_{4}(\frac{1}{16}) = -2\\
log_{4}(4^{-2}) = -2\\
-2 log_{4}(4) = -2\\
-2 = -2\\
\end{split} \]