Math Whiz Wanted!!! (for logarithmic equation)?

Question

Can somebody please solve this equation with details!! I'm desperate.
log(x+3) + log(x-6) =1

Answer 1

Never fear, a whiz is here.
Addition of logs is the analog to the multiplication of the log arguements (the stuff in these brackets), so you can rearrange to
log (x^2-3x-18)= 1
You can set both sides as exponents of the log base to get x^2-3x-18=10.
Then you can solve the quadratic for x (watch for an extraneous root-check your original eqtn)

Answer 2

one of the laws of a logarithm is that log (x*y)= log (x) + log (y).

So, the Log (x+3)+log(x-6) is the same as log [(x+3)(x-6)]
log (x^2-3x-18)=1

If you do not have a base, then the base is understood to be 10.

What a log is, is to solve for an exponient.

log(b)x = n means b^n = x where b is the base. So in this case you have 10^1=x^2-3x-18
0=x^2-3x-28
0=(x-7)(x+4)
x=7 or x=-4


(whenever a log is = 1, then the base is equal to the inside of the log)

Answer 3

Not a whiz, but here goes. You have not specified the base, so let us call it b; then raise both sides to base b:
b^{log(x+3) + log(x-6)} = b^1 = b
This can also be written:
{b^log(x+3)}{b^log(x-6)} = b
or, since b^log(a) = a,
(x+3)(x-6) = b
x^2 - 3x -18 = b
Whatever b is, you can determine x by means of the quadratic equation.

Answer 4

log(x + 3) + log(x - 6) = 1
log(x + 3)(x - 6) = 1 (since log a + log b = log (ab))
taking base as 10
log(x + 3)(x -6) = log 10 (since log 10 to the base 10 is 1)
taking out log
(x + 3)( x - 6) = 10
x^2 - 6x + 3x - 18 = 10
x^2 - 3x - 28 = 0
x^2 - 7x + 4x - 28 = 0
x(x - 7) + 4(x - 7) =0
(x - 7)(x + 4) = 0
x = 7 or x = - 4

Answer 5

log[(x+3)(x-6)]=1
i suppose it is lg (log10)
(x+3)(x-6)=10^1

I think you should know how to solve after this. multiply the two brackets then move 10 to left. solve quadratic equation.