Will give medal! Need help with logs!

find x when log (x^2 – 9) – 1 = log (x + 3)

Question

Will give medal! Need help with logs!

find x when   log (x^2 – 9) – 1 = log (x + 3)

anonymous · Answer

start with 
$$\log(x^2-9)-\log(x+3)=1$$

anonymous · Answer

then, using 
$$\log(A)-\log(B)=\log(\frac{A}{B})$$ rewrite the left hand side as 
$$\log(\frac{x^2-9}{x+3})=1$$

anonymous · Answer

right, I got that far but I dont know how the answer is x=13 =/

anonymous · Answer

I got x=4

anonymous · Answer

did you get to $$\log(x-3)=1$$?

anonymous · Answer

yes

anonymous · Answer

oh i see, i guess you did 
but it is not $x-3=1$ which would make $x=4$ it is 
$$\log(x-3)=1$$

anonymous · Answer

rewrite in equivalent exponential form as 
$$x-3=10$$

anonymous · Answer

huh, where did that 10 come from?

anonymous · Answer

good question 
$$\log(\spadesuit)=\diamondsuit \iff 10^{\diamondsuit}=\spadesuit$$

anonymous · Answer

base of $\log(whatever)$ is 10, i.e. it is $\log_{10}(x-3)=1$ and so $x-3=10^1$

anonymous · Answer

like if you had $\log(x-1)=2$ then $x-1=10^2$ etc

anonymous · Answer

so it would be|dw:1386641076055:dw|