# Solve for real x equation lgx+lg(9-2x)=1.

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We have to solve lg x + lg (9 - 2x) = 1 for x.

lg x + lg (9 - 2x) = 1

use the property of logarithms log a + log b = log (a*b)

=> lg[x(9 - 2x)] = 1

As log(a) a = 1, let us assume the base here is 10

=> x(9 - 2x) =10

=> 9x - 2x^2 = 10

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

=> 2x^2 - 5x - 4x + 10 = 0

=> x(2x - 5) - 2(2x - 5) = 0

=> (x - 2)(2x - 5) = 0

x = 2 and x = 5/2

For both these values of x, lg x and lg (9 - 2x) are defined.

**The solution of the equation is x = 2 and x = 5/2**

We'll impose the constraints of existence of the logarithms:

x>0

9-2x>0 => -2x>-9 => x < 9/2

The common interval of admissible values is (0 ; 9/2).

We'll solve the equation using the product rule of logarithms:

lg x*(9-2x) = 1

We'll take antilogarithm:

x*(9-2x) = 10

We'll remove the brackets:

9x - 2x^2 = 10

We'll move all terms to the left:

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

2x^2 - 9x + 10 = 0

We'll apply quadratic formula:

x1 = [9+sqrt(81 - 80)]/4

x1 = (9+1)/4

x1 = 5/2

x2 = 8/4

x2 = 2

**Since both x values belong to the interval (0 ; 9/2), we'll accept them as solutions: {2 ; 5/2}.**