Question

What values of a and b make the equation true? sqrt(648)=sqrt(2^a•3^b)

a) a=3, b=2
b) a=2, b=3
c) a=3, b=4
d) a=4, b=3

Answer 1

ANSWER

The value of a and b are:

$a=3,b=4$

EXPLANATION

The expression given to us is

$\sqrt{648} = \sqrt{ {2}^{a} \times {3}^{b} }$

We square both sides of the equation to obtain

$648= {2}^{a} \times {3}^{b}$

We now find the prime factorization of 648 to obtain,

${2}^{3} \times {3}^{4} = {2}^{a} \times {3}^{b}$

We now compare the exponents on both sides of the equation to obtain,

$a=3,b=4$

Therefore the correct option is C

Answer 2

The value of a = 4 and b = 3 make the equation true, the correct option is D.

Square root

The square root is defined as the value given the original value.

Given information

Equation; $\rm \sqrt{648} =\sqrt{(2^a\times 3^b)}$

The values of a and b are;

$\rm \sqrt{648} =\sqrt{(2^a\times 3^b)}\\\\\sqrt{2\times 2\times 2\times 2\times 3\times 3\times 3} =\sqrt{(2^a\times 3^b)}\\\\ \sqrt{2^4\times 3^3} =\sqrt{(2^a\times 3^b)}$

On comparing with the equation both sides;

The value of a is 4 and b is 3.

Hence, The value of a = 4 and b = 3 make the equation true, the correct option is D.

To know more about Square root click the link given below.

https://brainly.com/question/1969911