Ahhh! One last math problem to help me with. LOGS ARE HARD :(?

Question

http://online.math.uh.edu/webct/spring06/1310/pt5/images/pt5_f05_problem21__43.gif

By the way, thank you so much for helping me out (those who replies to my other boards) It means a lot, I have a final coming up!

Answer 1

Log(x + 2) + Log(x - 1) = Log(70)
As I explained before,
Log (a0 + Log (b) = Log(ab)
So,
Log(x + 2) + Log(x - 1) = Log [(x + 2)(x - 1)]
Log [(x + 2)(x - 1)] = Log(70)
(x+2)(x-1) = 70
x^2 - x + 2x - 2 = 70
x^2 + x = 72
x^2 + x = 72
x^2 + x - 72 = 0
x^2 - 8x + 9x - 72 = 0
x(x - 8) + 9(x - 8) = 0
(x + 9)(x - 8) = 0
Now, either x + 9 = 0 or x - 8 = 0
If we put x + 9 = 0, we get x = -9
If we put x - 8 = 0, we get x = 8
But when we substitute x = -9 in the original problem we get Log (-7) + Log (-10) = Log(70)
But Log of negative numbers if undefined.
So, x = 8 is the only solution.

Answer 2

Piece of cake. Remember that you are multiplying numbers when you add their logs. So you have the quadratic equation:
(x + 2)(x - 1) = 70
x^2 + x - 72 = 0
The roots are 8 and -9.
log(-7) and log(-10) are undefined, so the only real solution is 8.

Answer 3

What the first poster said!

since log (ab) = log a + log b you can change the left hand side.

Then exponentiate both sides and you'll get a quadratic in x, which you can solve.

(x+2)(x-1) = 70 .....

(don't use the quadratic formula, just recall that 9*8 = 72)

Answer 4

log [(x+2)(x-1)]=log 70
(x+2)(x-1)=70
x^2+x-2=70
x^2+x-72=0
(x+9)(x-8)=0
x={-9,8}
REJECT (-9) BECAUSE ONE CANNOT TAKE THE LOG OF A NEGATIVE NUMBER.

X=8

Answer 5

equivalent to:
(x+2)(x-1)=70
x^2+x-72=0
x=( -1 +/- sqrt(1+288) )/2
x=( -1 +/- 17)/2
x= -9, +8
drop -9 as it makes logs in original problem negative