# What is the answer for question for 6. a) ? http://postimg.org/image/xusvq33e3/

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You need to solve for x the logarithmic equation `log_3(x - 5) + log_3(x + 3) = 2` , hence, converting the sum of logarithms into the logarithm of product, using the logarithmic identity `log_a x + log_a y = log_a(x*y)` , yields:

`log_3(x - 5) + log_3(x + 3) = log_3 (x - 5)(x + 3) `

Replacing `log_3 (x - 5)(x + 3)` for `log_3(x - 5) + log_3(x + 3) ` yields:

`log_3 (x - 5)(x + 3) = 2 => (x - 5)(x + 3) = 3^2`

`x^2 + 3x - 5x - 15 - 9 = 0 => x^2 - 2x - 24 = 0`

Using quadratic formula, yields:

`x_(1,2) = (2+-sqrt(2^2 - 4*1*(-24)))/2`

`x_(1,2) = (2+-sqrt(4+96))/2`

`x_(1,2) = (2+-sqrt(100))/2`

`x_(1,2) = (2+-10)/2 => x_1 = 6; x_2 = -4`

Testing the values `x_1 = 6; x_2 = -4` in equation yields:

`log_3(6 - 5) + log_3(6 + 3) = 2 => log_3 1 + log_3 9 = 2`

`0 + log_3 3^2 = 2 => 2log_3 3 = 2 => 2*1 = 2` valid

`log_3(-4 - 5) + log_3(-4 + 3) = log_3(-9) + log_3(-1)` invalid

**Hence, evaluating the solution to the given equation yields x = 6.**