Here's the question you clicked on:
xBeEnchantedx
How do you solve: logbase5 (square root of x+3) = logbase5 (4)?
this means your insides have to be equal
\[\log_{5} (\sqrt{x+3})=\log_{5} (4)\] <- thats a better way to view the equation.
Since y=log_5(x) is a one to one function if you have f(a)=f(b) then this implies a=b
like basically i'm telling you that \[\sqrt{x+3}=4 \text{ now solve }\]
Ohhh thank you. I accidentally put the wrong exponent when trying to get rid of the square root, which through off my calculations. Thank you :D