Answer:
The solution is x = 2 and x = 4
Step-by-step explanation:
Let us revise some rules in logarithmic function:
Let us solve the question
∵ [tex]log_{2}(x)+log_{2}(6-x)=3[/tex]
→ By using the first rule above
∴ [tex]log_{2}(x)(6-x)=3[/tex]
→ Multiply x by (6 - x)
∵ x(6 - x) = x(6) - x(x) = 6x - x²
∴ [tex]log_{2}(6x-x^{2})=3[/tex]
→ Change the logarithmic function to exponential function using
the second rule above
∴ [tex]2^{3}=(6x-x^{2})[/tex]
∵ 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
