solving log equations?

Question

equation:
log x + log (x + 15) = 2

I know since the equation doesn't give a "base" it is assumed as 10. So the next step would be to change this log equation into an exponential equation but how do I solve for "x" when the equation is x^2 + 15x=100 ? maybe my exponential equation is wrong. Thanks for your help.

Answer 1

For log or ln, a + is the same as multiplying the insides. For example, log 10 + log 10 = 2 = log 100 = log (10*10).

So, in this case...

log x + log (x+15) = 2
log (x(x+15))=2
raise both sides to the power of 10

x^2+15x=100. So, you are right. Now, you have an equation with two possible solutions. x^2+15x-100=0, factor and solve for x.
(x+20)(x-5)=0; x=+5, -20

The answers that state you can't take the log of negative number are correct as far as they go. But, if log x + log (x+15)=log (x(x+15)), then you are not taking a log of a negative number. The original form does not work, but x can still be -20 and +5 which both give you log 100 = 2.

Answer 2

log x + log (x + 15) = 2
10^(log x + log (x + 15)) = 10^2
x(x+15) = 10^2
x^2 + 15 = 100
x^2 + 15 - 100 = 0
(x - 5)(x + 20) = 0
x = 5 or x =-20

Therefore x = 5, since -20 will not produce a real solution

Answer 3

log x + log (x + 15) = 2
condition: x>0
log{x(x+15)} = log(10^2)
x(x + 15) = 100
x2 + 15x - 100 = 0
(x + 20)(x - 5) = 0
Only x = 5 will do.

Th

Answer 4

x^2+15x-100=0
(x+20)(x-5)=0
x=-20 or 5

but we cant take the log of a negative number( we can but the result is imaginery), so the only correct solution is 5

Answer 5

(1) I take it that it is Log to base 5 of x is 3 and I need to find x x = 5^3 = 125. (2) Log to base 3 of (3x-2) is 2. Therefore, 3x-2=3^2 3x-2=9, 3x=11, therefore x=11/3 or 3.666666666..............