Question 1 of 10 Let log(b)A = 1; log(b)C = 3; log(b) D=4
What is the value of log(b) (a^5c^2/d^6)
A. 26
B. There isn't enough information to answer the question.
C. -13. D. 0

Log₆ (A⁵C² / D⁶) = –13

Step-by-step explanation:

From the question given above,

Log₆ A = 1

Log₆ C = 3

Log₆ D = 4

Log₆ (A⁵C² / D⁶) =?

Recall:

Log MN / U = Log M + Log N – Log U

Therefore,

Log₆ (A⁵C² / D⁶) = Log₆ A⁵ + Log₆ C² – Log₆ D⁶

Recall:

Log Mⁿ = nLog M

Thus,

Log₆ A⁵ + Log₆ C² – Log₆ D⁶

= 5Log₆ A + 2Log₆ C – 6Log₆ D

Log₆ A = 1

Log₆ C = 3

Log₆ D = 4

= 5(1) + 2(3) – 6(4)

= 5 + 6 – 24

= 11 – 24

= –13

Therefore,

Log₆ (A⁵C² / D⁶) = –13

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

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: