SOLUTION: What is the solution of the equation? 4(3-x)^4/3 - 5 = 59 A. –5, 11 B. 5 C. 11 D. –11

Add 5 to both sides 

    

Divide both sides by 4

    

Now we use the principle of fraction exponents to radicals 

    

We use the principle of even roots, in this case 4th roots. When taking
even roots we must use ± on the right side. Now since 16 = 2·2·2·2 = 24,
then the 4th root of 16 is 2. So we have:

    

Next to get rid of the cube root we cube both sides:

    

    3-x = ±8

If we use the +

    3-x = 8
     -x = 5
      x = -5

If we use the -

    3-x = -8
     -x = -11
      x = 11

So we get two solutions -5, and 11.  We check to see if either one
is extraneous:

Checking x = -5,









Write 8 as 2³



Multiply the exponents to remove the parentheses:

4(24) - 5 = 59
4(16) - 5 = 59
64 - 5 - 59
    59 = 59

So -5 is a valid solution.

Checking x = 11:










Write 8 as 2³



Multiply the exponents to remove the parentheses:

4((-2)4) - 5 = 59
4(16) - 5 = 59
64 - 5 - 59
    59 = 59

So 11 is also a valid solution.

Solutions:  -5 and 11

Edwin

 

Hi,

4(3-x)^4/3 - 5 = 59

(3-x)^4/3 = 16



   (3-x) = 16^3/4 = 

  (3-x) = 8

    -5 = x