Question

Will give medal! Need help with logs!

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

I got x=4

Answer 5

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

Answer 6

yes

Answer 7

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

Answer 8

rewrite in equivalent exponential form as
\[x-3=10\]

Answer 9

huh, where did that 10 come from?

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

so it would be