## College Algebra (6th Edition)

$x=21$
From the definition of the logarithmic function on page 456, we know that $y=log_{b}x$ is equivalent to $b^{y}=x$ (for $x\gt0$, $b\gt0$, and $b\ne1$). Therefore, $log_{5}(x+4)=2$ is equivalent to $5^{2}=(x+4)$ in logarithmic form. We know that $5^{2}=25$, so we can set $x+4=25$. Subtract 4 from both sides. $x=21$