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Solve for x. lg(x-3)+lg(x+6)=lg2+lg5?

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luvgoj | Student, Undergraduate | Honors

Posted September 8, 2013 at 3:12 PM via web

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Solve for x. lg(x-3)+lg(x+6)=lg2+lg5?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 8, 2013 at 3:21 PM (Answer #2)

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You need to use the properties of logarithms and to convert the summation of logarithms into the logarithm of product, such that:

`lg(x - 3)(x + 6) = lg(2*5)`

`lg (x^2 + 6x - 3x - 18) = lg 10`

Equating the arguments of logarithms yields:

`x^2 + 3x - 18 = 10 => x^2 + 3x - 28 = 0`

Using quadratic formula yields:

`x_(1,2) = (-3+-sqrt(9 + 4*28))/2 => x_(1,2) = (-3+-sqrt121)/2`

`x_(1,2) = (-3+-11)/2 => x_1 = 4 ; x_2 = -7`

Testing the values `x_1 = 4 ; x_2 = -7` in equation yields:

`x_1 = 4 => lg(4 - 3)(4 + 6) = lg(2*5) => lg(1*10) = lg 10 => lg 10 = lg 10 => 1 = 1` valid

`x_2 = -7 => lg(-7 - 3) + lg (-7 + 6) = lg(2*5)`

`lg(-10) + lg(-1) = lg 10`

Since the arguments of logarithms cannot be negative, hence, the value `x = -7` is invalid.

Hence, evaluating the solution to the given equation, yields `x = 4.`

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