log( x + 12 ) + log (x - 3) = 2 solve for x.?

Question

please solve for x. an explanation would be nice. Thanks

Answer 1

log( x + 12 ) + log (x - 3) = 2
log[(x+12)(x-3)] =2
assuming base 10:
10^2 = (x+12) (x-3)
100 = x^2 + 9x - 36
x^2 + 9x - 136 = 0
(x - 8) (x + 17) = 0
x = 8, x=-17
-17 is extraneous (check)

so x=8

Answer 2

We would need to now the base of the logarithm, if it's base ten (standard for notation log) the solutions will be obtained by:
first we combine the logarithms to get log((x+12)(x-3))=2 then we exponentiate 10 by both sides of the equation, thus preserving the equality. 10^(log((x+12)(x-3))=10^2 but this is:
x^2+9x-36=100 (left side is by definition of logarithm) wich becomes x^2+9x-136=0 wich can be solved by the general formula, giving x=-17 or x=8 but x=-17 gives us a non real result on log(x+12) so we take x=8 as our solution.
If the base were another the number the procedure is analogus.

Answer 3

Two assumptions: (1) you know how to solve a quadratic equation and (2) the log is to base 10. To solve this you have to remember a basic rule of logarithms, namely, log(A*B) = log(A) + log(B), that is, the log of the product is the sum of the logs. So,

log(x + 12) + log(x - 3) = log((x+12)*(x-3)) = log(x^2 +9x - 36) = 2. This implies, upon exponentiating both sides, x^2 + 9x - 36 = 10^2 = 100 or x^2 + 9x -136 = 0. From here I will let you solve for x (hint, x = -17 or 8).

Math Rules!

Answer 4

Assume logs are to base ten.
Let "log" mean log base ten in the following:-
log (x + 12).(x - 3) = 2
(x + 12).(x - 3) = 10²
x² + 9x - 136 = 0
(x - 8).(x + 17) = 0
x = 8, x = - 17
Accepting positive value, x = 8

Answer 5

well, this is quite straight forward.
remember that the sum of the logs, is the log of the products.. which then gives us,
log(x+12)(x-3)=2
which follows to
(x+12)(x-3)=10^2 where here the base is 10..

now, multiply and regroup to get

x^2+9x -136=0

on factoring, we end up with (x+17)(x-8)=0.. which gives
x=-17, and x=8.. however, if we substitute back, x=-17 gives a false answer, since you can't take the log of a negative, so the only true solution is then x=8..

hope this helps for future problems..

Answer 6

log( x + 12 ) + log (x - 3) = 2
log[(x+12)(x-3)]=log 10^2
(x+12)(x-3)=10^2
x^2+9x-36-100=0
x^2+9x-136=0
(x-8)(x+17)=0
x=8 or x=-17(ignore)

so final answer is x=8#

Answer 7

log(x + 12) + log(x - 3) = 2
log((x + 12)(x - 3)) = 2
log(x^2 - 3x + 12x - 36) = 2
log(x^2 + 9x - 36) = 2
x^2 + 9x - 36 = 10^2
x^2 + 9x - 36 = 100
x^2 + 9x - 136 = 0
(x + 17)(x - 8) = 0
x = -17 or 8

since you can't have log(-20)

x = 8

Answer 8

log( x + 12 ) + log (x - 3) = 2
log[(x+12)(x-3)]=2
(x+12)(x-3)=antilog 2
(x+12)(x-3)=100
x^2+9x-36=100
x^2+9x-136=0
x^2+17x-8x-136=0
x(x+17)-8(x+17)=0
(x+17)(x-8)=0
Either x+17=0 --> x= -17
or x-8=0 --> x=8
Ans. x=8, -17

Answer 9

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