# What is x for lg(x)+lg(x+1)=(1+lg9) to exist?

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### 4 Answers

lg(x)+ lg(x+1) = (1+lg9)

We will use the logarithm properties to solve for x.

First, we know that lg a + lg b = lg (ab)

==> lg x*(x+1) = 1+ lg9

==> lg (x^2+x) = 1+ lg9

Now we know that lg10 = 1

==> lg (x^2 +x) = lg 10 + lg9

==> lg (x^2 + x) = lg (10*9)

==> lg (x^2 +x) = lg (90)

Now that the logs are equal, then the bases are equal.

==> x^2 +x = 90

==> x^2 + x - 90 = 0

Now we will factor;

==> (x +10) (x-9) = 0

==> x1 = 9

==> x2= -10 ( But this answer is not valid because the log is not defined for negative numbers.

**Then the only solution is x=9**

Using the product rule for logarithms.

lg x + lg(x +1) = lg[ x(x+1)]

This is equal to 1 + lg 9

=> lg[ x(x+1)] = 1 + lg 9

=> lg[ x(x+1)] - lg 9 = 1

Now 1 = lg 10

=> lg [ x(x+1)]/9] = lg 10

=> x(x+1)/9 = 10

=> x^2 + x = 90

=> x^2 + x - 90 =0

=> x^2 + 10x - 9x - 90 =0

=> x( x + 10) - 9(x +10) =0

=> (x - 9)(x +10) = 0

x can be 9 or -10

But the log of negative number is not defined, so x = -10 can be eliminated

**Therefore x = 9.**

What is x for lg(x)+lg(x+1)=(1+lg9) to exist?

lgx+lg(x+1) = 1+lg9.

lgx(x+1) = lg10+lg9 , as lga+lgb = lgab.

lgx(x+1) = log90.

We take antilog:x(x+1) = 90.

x^2+x-90 = 0.

(x+10)(x-9) = 0 gives the valid solution.

x+10 = gives x= -10, for which lg (-10) does not exist.

x+10 = 0, or x-9 = 0.

So x-9 = 0 gives x=9.

We'll impose the constraints of existence of logarithms:

x > 0

x + 1 > 0

x > -1

The interval of admissible values for x is (o ; +infinite).

We'll re-write the equation and we'll put lg 10 instead of value of 1:

lg(x+1) + lgx = 1 + lg9 is changing into lg(x+1) + lgx = lg 10 + lg9

Since the bases are matching, we'll use the product rule of logarithms both sides:

lg x(x+1) = lg 90

Since the bases are matching, we'll use one to one property:

x(x+1) = 90

We'll remove the brackets:

x^2 + x - 90 = 0

We'll apply the quadratic formula:

x1 = [-1+sqrt(1+360)]/2

x1 = (-1+19)/2

x1 = 9

x2 = (-1-19)/2

x2 = -10

**We'll reject the second solution, because it's negative. We'll keep the valid solution, x = 9.**