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

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

∵

→ By using the first rule above

∴

→ Multiply x by (6 - x)

∵ x(6 - x) = x(6) - x(x) = 6x - x²

∴

→ Change the logarithmic function to exponential function using

   the second rule above

∴

∵ 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

24

step-by-step explanation:

i assuming that 2 sides are 6 the other 2 sides are 4 cm so 6 times 4 is 24 : )

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2