Question

Log2 x + log2 (6-x)=3 solve for x algebraically

Answer 1

The solution is x = 2 and x = 4

Explanation:

Let us revise some rules in logarithmic function:

• 
  `log_(a)(x)+log_(a)(y)=log_(a)(xy)`

• The logarithmic function is the inverse of the exponential function

• 
  `log_(b)(c)=a` is the inverse of
  `b^(a)=c`

Let us solve the question

∵
`log_(2)(x)+log_(2)(6-x)=3`

→ By using the first rule above

∴
`log_(2)(x)(6-x)=3`

→ Multiply x by (6 - x)

∵ x(6 - x) = x(6) - x(x) = 6x - x²

∴
`log_(2)(6x-x^(2))=3`

→ Change the logarithmic function to exponential function using

the second rule above

∴
`2^(3)=(6x-x^(2))`

∵ 2³ = 8

∴ 8 = 6x - x²

→ Add x² to both sides

∴ x² + 8 = 6x

→ Subtract 6x from both sides

∴ x² - 6x + 8 = 0

→ Factorize it into 2 brackets

∴ (x - 2)(x - 4) = 0

→ Equate each factor by 0 to find the values of x

∵ x - 2 = 0

→ Add 2 to both sides

∴ x = 2

∵ x - 4 = 0

→ Add 4 to both sides

∴ x = 4

∴ The values of x are 2 and 4

∴ The solution of the equation is x = 2 and x = 4