Question 258790
Solve for x:
{{{(x-3)^(2/5) = (4x)^(1/5)}}} Raise both sides to the 5th power.
{{{((x-3)^(2/5))^5 = ((4x)^(1/5))^5}}} Multiply the exponents on the inside of the parentheses by those on the outside.
{{{(x-3)^2 = (4x)}}} Simplify.
{{{x^2-6x+9 = 4x}}} Subtract 4x from both sides.
{{{x^2-10x+9 = 0}}} Factor this quadratic equation.
{{{(x-1)(x-9) = 0}}} Apply the zero product rule.
{{{x-1 = 0}}} or {{{x-9 = 0}}} so...
{{{highlight(x = 1)}}} or {{{highlight(x = 9)}}}
Caution! Check the solutions by substituting into the given equation to discover any possible "extraneous" solutions..
{{{(x-3)^(2/5) = (4x)^(1/5)}}} Substitute x = 1.
{{{(1-3)^(2/5) = (4*1)^(1/5)}}} Evaluate using a calculator.
{{{(-2)^(2/5) = (4)^(1/5)}}}
{{{1.3195079 = 1.3195079}}} OK for x=1.
{{{(x-3)^(2/5) = (4x)^(1/5)}}} Substitute x = 9.
{{{(9-3)^(2/5) =(4*9)^(1/5)}}} Evaluate.
{{{6^(2/5) = 36^(1/5)}}}
{{{2.0476725 = 2.0476725}}} OK for x = 9
Both solutions are valid!