# Solve the logarithmic equation, express solutions in exact form only. log₂(x+8)+log₂(x-7)=4+log₂(x-3)PLEASE SHOW ALL WORK!!!!!

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### 1 Answer

You need to bring all logarithms to the left side and to use the properties of logarithms having bases alike such that:

`log_2 (x+8) + log_2 (x-7) - log_2 (x-3) = 4`

You need to remember that the sum of two logarithms having like bases may be transformed in the logarithm of the product such that:

`log_2 (x+8) + log_2 (x-7) = log_2 (x+8) *(x-7)`

You need to remember that the difference of two logarithms having like bases may be transformed in the logarithm of the quotient such that:

`log_2 (x+8) *(x-7) - log_2 (x-3) = log_2 ((x+8) *(x-7))/(x-3)`

Writing the transformed form of equation yields:

`log_2 ((x+8) *(x-7))/(x-3) = 4 =gt ((x+8) *(x-7))/(x-3) = 2^4`

Multiplying by x - 3 both sides yields:

`((x+8) *(x-7)) = 16(x-3)`

Opening the brackets yields:

`x^2 - 7x + 8x - 56 = 16x - 48`

You need to move all terms to the left side such that:

`x^2 - 15x - 8 = 0`

Using the quadratic formula yields:

`x_(1,2) = (15+-sqrt(225 + 32))/2 =gt x_1~~ (15+16.07)/2`

`x_1~~15.51`

`` `x_2~~-0.535`

Notice that for each logarithm to exist, the values of x need to be at most 7, hence the second value of x fails.

**Hence, the solution to the equation is `x_1~~15.51` .**