Can you help me solve: log10x + log10 (x + 3) = log10(x - 1). (The 10's are subscripts) Thanks!?

Question

I'm not sure how to select best answer, but I want to thank you all for your help. I go to x^2 + 2x = 1, but I couldn't figure out what to do next. Thanks so much, I couldn't find an example like this in my book. Thanks again!

Answer 1

The fact that these are all of base 10 is extremely helpful, because now you can eliminate the logs from each term. You want to remember your log rules though. Think about logs as powers, when you multiply bases you add the powers together. So since these powers are being added, I know that I can rewrite this as:

x times (x + 3) = x - 1
x(x+3) = x-1
x^2 + 3x = x - 1
x^2 + 2x + 1 = 0
(x+1)(x+1) = 0

So x = -1

You must always remember to check your answer though, since you can't take the log of a negative number. Since this causes a negative number to occur in two of the terms, we have to reject this answer, which means THERE IS NO SOLUTION to this problem.

Hope this helped!

Answer 2

These are all base 10, so we use the following rules:

10 ^ (log10x) = x
10^(a+b) = (10^a)*(10^b)

Applying these rules:

10^[log10x + log10(x+3)] = 10^[log10(x-1]
(10^[log10x])*(10^[log10(x+3)]) = 10^[log10(x-1)]
x(x+3)=x-1
x^2 + 3x = x - 1
x^2 + 2x + 1 = 0
(x+1)(x+1) = 0
x = -1

You can't take the log of a negative number, so there is no solution to this problem.

Answer 3

If log10x+ log10(3x+a million) = a million it rather is comparable to log10 (x*(3x+a million)) = a million Taking antilogs from the two factors = x(3x+a million) = 10^a million x(3x+a million) - 10 = 0 3x^2 +x - 10 = 0 This factorises to (x+2)(x- 5/3)= 0 verify x^2 - (5/3)x +2x - 10/3 x^2 - (5/3)x + (6/3)x - 10/3 x^2 + (a million/3)x - 10/3 = 0 3x^2 + x -10 = 0 consequently x= -2 or + 5/3.

Answer 4

I've left my algebra long time ago. But hope this is correct.

log10 (x) + log10 (x+3) = log10 (x-1)
log10 (x+x+3-x+1) = 0
log10 (-x+4) = 0
log10 -x + log10 4 = 0
-log10 x = -log10 4
x=4

Answer 5

First all numbers must be >0 so x-1>0 x>1 covers all
x(x+3)= x-1
x^2+2x+1=0 x=(-2+-sqrt(4-4))/2=-1 which can´t be as negative numbers don´t have log

Answer 6

they all the same log base....therefore:
x(x+3)=(x-1)
x^2+3x=x-1
x^2+2x+1=0
(x+1)(x+1)=0
x=-1

Answer 7

no