log(x+1) = log10+log(x-1).
To find x, we use the logarithms rules:
loga +logb = logab.
Therefore we can write the expression on the right ,log10+log(x-1) as log10*(x-1).
So now we can write the given equations as:
log(x+1) = log10(x-1)
We take antilog:
x+1 = 10(x-1).
x+1 = 10x-10
1+10 = 10x-x
11 = 9x.
11/9 = x.
Or x= 11/9
For logarithms, we use the formula log (a*b) = log a + log b.
Here we have lg(x+1) = lg10 + lg(x-1)
=> lg(x+1) = lg10 + lg(x-1)
=> lg(x+1) = lg [10*(x-1)]
take the antilog of both the sides
=> x+1 = 10*(x-1)
=> x+1 = 10x - 10
take the numbers to one side and the terms with x to the other
=> 10x - x = 10 +1
=> 9x = 11
=> x = 11/9
Also we see that 11/9 + 1 and 11/9 -1 are positive, therefore their logarithms are defined
So the required value of x is 11/9
First, we'll impose the constraints of existence of logarithms:
The interval of admissible values for x is (1,+inf).
Now, we'll solve the equation:
lg(x+1) = lg10 + 2lg(x-1)
We'll apply the power rule of logarithms for the term 2lg(x-1):
2lg(x-1) = lg (x-1)^2
We'll re-write the equation:
lg(x+1) = lg10 + lg(x-1)^2
Because the bases of the logarithms from the right side are matching, we'll apply the product rule of logarithms:
lg a + lg b = lg a*b
We'll put a = 10 and b = (x-1)^2
lg(x+1) = lg10(x-1)^2
Since the bases are matching, we'll apply one to one property:
x+1 = 10(x-1)^2
We'll expand the square from the right side:
x+1 = 10(x^2 - 2x + 1)
We'll remove the brackets:
x+1 = 10x^2 - 20x + 10
We'll subtract x+1 both sides and we'll apply symmetric property:
10x^2 - 20x + 10 - x - 1 = 0
10x^2 - 21x + 9 = 0
We'll apply the quadratic formula:
x1 = [21+sqrt(441-360)]/20
x1 = (21+9)/20
x1 = 30/20
x1 = 3/2 =1.5 > 1
x2 = (21-9)/20
x2 = 12/20
x2 = 3/5 = 0.6 < 1
Since the second root doesn't belong to the interval of admissible value, this one will be rejected and the equation will have only one root, x = 1.5.