SOLUTION: I am trying to solve log(x-1)+log(x+2)=1, I have the answer, which is 3, but am looking to see the problem worked. I appreciate the help. Thank you.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I am trying to solve log(x-1)+log(x+2)=1, I have the answer, which is 3, but am looking to see the problem worked. I appreciate the help. Thank you.       Log On


   

Question 289559: I am trying to solve log(x-1)+log(x+2)=1, I have the answer, which is 3, but am looking to see the problem worked. I appreciate the help. Thank you.

Found 2 solutions by jim_thompson5910, CharlesG2:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Note: All logs are assumed to be base 10.


Hint: Use the idea that log%28%28x%29%29%2Blog%28%28y%29%29=log%28%28xy%29%29 to get log%28%28%28x-1%29%28x%2B2%29%29%29=1. Now convert into exponential form to get %28x-1%29%28x%2B2%29=10%5E1 and simplify to get %28x-1%29%28x%2B2%29=10. I'll let you take it from here.

Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
I am trying to solve log(x-1)+log(x+2)=1, I have the answer, which is 3, but am looking to see the problem worked. I appreciate the help. Thank you.
logb (mn) = logb (m) + logb (n) by rules of logarithms
just plain log is usually assumed to be log base 10 or log10
log(x-1) + log(x+2) = log[(x-1)(x+2)] = 1
log(x^2 + x - 2) = 1
in logb y = x, b^x = y by rules of logarithms
10^1 = x^2 + x - 2
10 = x^2 + x - 2
0 = x^2 + x - 12
0 = (x + 4)(x - 3) (by FOIL this equals above)
x = -4 or x = 3 (check which x works as the answer)
check with x = -4: log(x - 1) = log(-5) not doable in realm of real numbers
log(x + 2) = log(-2) also not doable in realm of real numbers
check with x = 3: log(x - 1) = log(2), log(x + 2) = log(5), these 2 are solvable
within realm of real numbers
so answer is x=3