Log(x+1)=Log3-log(2x-1)?

Question

can someone explain to me how to solve this

Answer 1

You can combine the right side terms using the
property log(a/b) = log a - log b in reverse.

It is log(3/(2x-1))

So now you have log(x+1) = log(3/(2x-1)

Since the logs are equal, so must be the things you took the log of.
So x+1 = 3/(2x-1)

Cross multiply: (x+1)(2x-1) = 3

2x^2 + x - 1 = 3

2x^2 + x - 4 = 0

Solve for x. Check that when you plug it in, you don't end up with the log of a negative number.

Answer 2

Rules of logarithms state that log a - log b = log a/b, so

log (x+1) = log 3 - log (2x-1) becomes

log (x+1) = log 3/(2x-1) Also, when log a = log b, a = b, so

x+1 = 3/(2x-1)

(x+1)(2x-1) = 3
2x^2-x+2x-1=3
2x^2+x-1=3
2x^2+x-4=0
Plug into quadratic formula, and

x= (-1+ or - square root of 33)/4

Answer 3

x has to be > 1/2
Log(x+1)=Log3-log(2x-1)
<=> log(x+1) + log(2x-1) = log3
<=> log[(x+1)*(2x-1)] = log3
<=> 2x^2 + x - 1 = 3
<=> 2x^2 + x - 4 =0
delta = 1 + 32 = 33
x = (-1+square of 33)/4 (<1/2)
x = (-1-square of 33)/4 (< 1/2)
then this equation has no root

Answer 4

log(x + 1) = log(3) - log(2x - 1)
log(x + 1) = log(3/(2x - 1))
x + 1 = (3/(2x - 1))
(2x - 1)(x + 1) = 3
2x^2 + 2x - x - 1 = 3
2x^2 + x - 1 = 3
2x^2 + x - 4 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-1 ± sqrt(1^2 - 4(2)(-4)))/(2(2))
x = (-1 ± sqrt(1 + 32))/4
x = (-1 ± sqrt(33))/4

x = (1/4)(-1 ± sqrt(33))

Since you can't have a negative log

ANS :

x = (1/4)(-1 + sqrt(33))