SOLUTION: Please solve : log (k+2) + log (k-1) = 1

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Question 936216: Please solve : log (k+2) + log (k-1) = 1
Found 2 solutions by ewatrrr, srinivas.g:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
log (k+2) + log (k-1) = 1 Note: 10^1 = 10
(k+2)(k-1) = 10
k^2 + k - 12 = 0
{k+4)(k-3)= 0
k = -4, 3
.......
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Answer by srinivas.g(540) About Me  (Show Source):
You can put this solution on YOUR website!
log (k+2) + log (k-1) = 1 log 10 =1
log (k+2) + log (k-1) = log 10
formula : log a+log b = log(a*b)
log (k+2)*(k-1)=log 10
(k+2)(k-1)=10
k(k-1)+2(k-1)=10
+k%5E2%2Bk%2A%28-1%29%2B2%2Ak%2B2%2A%28-1%29++=10
+k%5E2-k%2B2k-2+=10
+k%5E2%2Bk-2=10
move 10 to the left
+k%5E2%2Bk-2-10=0
+k%5E2%2Bk-12=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ak%5E2%2Bbk%2Bc=0 (in our case 1k%5E2%2B1k%2B-12+=+0) has the following solutons:

k%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A-12=49.

Discriminant d=49 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+49+%29%29%2F2%5Ca.

k%5B1%5D+=+%28-%281%29%2Bsqrt%28+49+%29%29%2F2%5C1+=+3
k%5B2%5D+=+%28-%281%29-sqrt%28+49+%29%29%2F2%5C1+=+-4

Quadratic expression 1k%5E2%2B1k%2B-12 can be factored:
1k%5E2%2B1k%2B-12+=+1%28k-3%29%2A%28k--4%29
Again, the answer is: 3, -4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-12+%29