# Solve for xlg(x+1)-lg9=1-lgx

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### 2 Answers

The equation given is lg(x+1) - lg 9 = 1 - lg x

I assume the base of the logarithm is 10. Use the property that lg a - lg b = lg a/b and 1 = lg 10

lg(x+1) - lg 9 = 1 - lg x

=> lg(x+1) - lg 9 = lg 10 - lg x

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

(x + 1)/9 = 10/x

=> x^2 + x = 90

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

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

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

x = 9 and x = -10

As log of negative numbers is not defined we eliminate x = -10

**The solution of the equation is x = 9**

We'll re-write the equation, moving the terms in x to the left side and the terms without x, to the right side.

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

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

We'll re-write 1 as lg 10

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 rule:

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 solution x = 9.**