Question

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

Answer 1

Answer:

The solution is x = 2 and x = 4

Step-by-step 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