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

Question 936216: Please solve : log (k+2) + log (k-1) = 1
Found 2 solutions by ewatrrr, srinivas.g:
Answer by ewatrrr(24785) (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) (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

move 10 to the left

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=49 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 3, -4. Here's your graph:

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