What is x if lg(x+1) = lg10 + 2lg(x-1) ?

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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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

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll impose the constraints of existence of logarithms:

x+1>0

x>-1

x-1>0

x>1

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.

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