Expand the logarithmic expression. log5(7)a^5 a. log57 • 5log5a b. log57 + 5log5a c. 7log5a^5 d. log57 – 5log5a

The answer for the exercise shown above is the option B, which is: 
 B. log5(7)+5log5(a)
 The explanation of the problem is shown below:
 1. You have the following expression given in the problem above:
 log5(7)(a^5)
 2. To expand it, you must use the logaritms properties, as following:
 log5(7)(a^5)
 log5(7)+log5(a^5)
 log5(7)+5log5(a)
 3. Therefore, as you can see, the answer is the option B.
 
 
 

2x^2 -2x -12 = 02(x -3)(x +2) = 0see below for explanation

step-by-step explanation:

equation

you are asked to write the equation

  expression = 0

using the given expression. when you do that, you have

  2x^2-2x-12 = 0

factoring

you can see that each of the coefficients is even, so a factor of 2 can be removed immediately.

  2(x^2 -x -6) = 0

if you multiply the binomials (x+a)(x+b), you get

  (x +a)(x +b) = x^2 +(a+b)x +ab

this tells you the coefficient of x is the sum of two factors of the constant term. here, that means -1 is the sum of two factors of -6. we want to know those factors so we can factor the trinomial.

  -6 = -1×6 = -2×3 = -3×2 = -6×1

the factor pair that sums to -1 is {-3, 2}, so our factorization is

  2(x -3)(x +2) = 0

solutions

the solutions to this equation are the values of x that make the product zero. a product is zero only when only of the factors of the product is zero. here, that means

  x - 3 = 0

  x = 3 . . . add 3 to both sides of the equation

or

  x +2 = 0

  x = -2 . . . add -2 to both sides of the equation

the solutions to the equation are x=3 or x=-2. both of these values of x satisfy the equation—they make it be true.

What’s the improper fraction?