2^a = 5^b = 20^c, express c in terms of and b

c=ab/(2b+a)

Step-by-step explanation:

20^c = 4^c  *  5^c

20^c=  (2^2)^c   * 5^c

20^c= (2 )^(2c)   * ( 5 )^c

Raise both sides to power a

20^(ca)=(2^a)^(2c)  * (5  )^(ac)

20^(ca)=(20^c)^(2c) * (5  )^(ac)

Raise both sides to power b

20^(cab)=(20)^(2c^2b)*(5^b)^(ac)

20^(cab)=20^(2c^2b) * (20^c)^(ac)

20^(cab)=20^(2c^2b) * 20^(ac^2)

Rewriting using law of exponents on right hand side

20^(cab)=20^(2c^2b+ac^2)

Now bases are same so that means the exponents have to be the same, that is we have:

cab=2c^2b+ac^2

assuming c is not 0, divide by c on both sides

ab=2cb+ac

Factor the right hand side

ab=c(2b+a)

Divide both sides by (2b+a)

c=ab/(2b+a)

answer: the correct option is d.

explanation:

it is given that the triangle j'k'l' shown on the grid below is a dilation of triangle jkl using the origin as the center of dilation.

from the given graph it is noticed that k(4,8) and k'(2,4). let origin is defined by o and scale factor is defined by k, then

    (1)

distance formula,

put these values in equation (1).

therefore scale factor is     and option d is correct.