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

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Answer:

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

User Brian Tarbox
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