Logarithms

[carebear]

Question:

log base 10 of (x + 2) + log base 10 of (x -1) = 1

I think the solution is:

log (x+2)(x-1) = 1
log (x^2 + x - 2) = 1 (step 2)
10^1 = x^2 + x - 2
0=x^2 + x - 12
0=(x+4)(x-3)
x = -4 and 3 but I believe I would reject -4 as it would make the logarithm in the original question undefined.

.....or if for example I were to substitute -4 into the equation in step 2, the logarithm would not be undefined because the negative value was squared......Am I right to check my answers with the original question?

 

[Deleted member 4993]

Question:

log base 10 of (x + 2) + log base 10 of (x -1) = 1

I think the solution is:

log (x+2)(x-1) = 1
log (x^2 + x - 2) = 1 (step 2)
10^1 = x^2 + x - 2
0=x^2 + x - 12
0=(x+4)(x-3)
x = -4 and 3 but I believe I would reject -4 as it would make the logarithm in the original question undefined.

.....or if for example I were to substitute -4 into the equation in step 2, the logarithm would not be undefined because the negative value was squared......Am I right to check my answers with the original question?

Your work is correct. Specially I like your "checking answers" - with the original equation. That is always required when you are trying to solve an equation.