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