# Solve for x: lg9 + 1 = lg(x+1) + lgx.Solve for x: lg9 + 1 = lg(x+1) + lgx.

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### 4 Answers

log 9 + 1 = log(x+1) + log x

We know that log 10 = 1

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

Also, we know that: log a + log b = log a*b

==> log 9*10 = log (x+1)*x

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

==> 90 = x^2 + x

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

Now let us factor:

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

==> x1= 9

==> x2= -10 (impossible because log x is not defined for negative values.

Then the answer is:

x = 9

lg9+1 =lg(x+1)+lgx.To solve for x.

Solution:

By property of logarithms, lga+l lgb = ab. So we can rewrite the given equation as:

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

lg9 + lg10 = lg(x^2+x), as log 10 = 1.

lg90 = lg(x^2+x) . Take antilogarithms.

90 = x^2+x .

x^2+x-90 = 0

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

x+10 = 0 Or x-9 = 0

x =-10 or x = 9.

We have to solve for x in lg9+1=lg(x+1)+lgx.

The base of the log can be taken to be anything as long as it is the same throighout the equation. Let's take it as 10

Now log 10 to the base 10 is 1 or 1= lg10

So rewrite the equation as lg9+lg10=lg(x+1)+lgx

The sum of the log of two numbers is the log of the product of the two numbers.

This means lg9+lg10=lg9*10=lg90=lg((x+1)*x)

We also know that if lg a= lg b it implies a=b.

So we get 90=(x+1)*x=x^2+x. Solving we get the quadratic equation : x^2+x-90=0

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

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

(x-9)(x+10)=0 or x can be 9 or -10

As the log of a negative number is not defined, -10 is not a possible alternative, so the right answer is **x=9**

We'll impose the constraints of existence of logarithms:

x>0

and

x+1>0

x>-1

The common interval of admissible values for x is (0 , +inf.).

Now, we'll solve the equation, applying th product rule of the logarithms:

lg9 + lg 10 = lg(x+1) + lgx

lg 9*10 = lg[x*(x+1)]

Because the bases of logarithms are matching, we'll use the one to one property:

x*(x+1)=9*10

We'll remove the brackets and we'll move all terms to one side:

x^2 +x -90=0

We'll apply the quadratic formula:

x1=[-1+ sqrt(1+4*90)]/2=(-1+19)/2=9 > 0

x2=(-1-19)/2=-10 < 0

**We notice that the second solution is not in the interval of admissible values for x, so the equation will have just a single solution, x = 9.**