Solve for x (5x-4)^(2/3)=x^(1/3)

Solution:

Step:1

Raise both sides of the equation to the power of 3 to clear the fractional exponents.

$((5x-4)^{\frac{2}{3}})^{3} = (x^{\frac{1}{3}})^{3}$

Step:2

Apply the power of a power rule to the left side of the equation.

Step:2.1

Utilize the rule $(a^{m})^{n} = a^{mn}$.

$((5x-4)^{\frac{2}{3} \cdot 3}) = (x^{\frac{1}{3}})^{3}$

Step:2.2

Simplify by canceling out the 3s.

Step:2.2.1

Perform the cancellation.

$((5x-4)^{\frac{2}{\cancel{3}} \cdot \cancel{3}}) = (x^{\frac{1}{3}})^{3}$

Step:2.2.2

Rewrite the simplified expression.

$(5x-4)^{2} = (x^{\frac{1}{3}})^{3}$

Step:3

Simplify the right side of the equation.

Step:3.1

Apply the power of a power rule to the right side.

Step:3.1.1

Use the rule $(a^{m})^{n} = a^{mn}$.

$(5x-4)^{2} = x^{\frac{1}{3} \cdot 3}$

Step:3.1.2

Simplify by canceling out the 3s.

Step:3.1.2.1

Perform the cancellation.

$(5x-4)^{2} = x^{\frac{1}{\cancel{3}} \cdot \cancel{3}}$

Step:3.1.2.2

Rewrite the simplified expression.

$(5x-4)^{2} = x^{1}$

Step:3.2

Simplify further.

$(5x-4)^{2} = x$

Step:4

Isolate the terms with x on one side.

Step:4.1

Subtract x from both sides.

$(5x-4)^{2} - x = 0$

Step:4.2

Expand and simplify.

Step:4.2.1

Express $(5x-4)^{2}$ as $(5x-4)(5x-4)$.

$(5x-4)(5x-4) - x = 0$

Step:4.2.2

Use the FOIL method to expand.

Step:4.2.2.1

Distribute each term.

$5x(5x-4) - 4(5x-4) - x = 0$

Step:4.2.2.2

Continue distributing.

$5x(5x) + 5x(-4) - 4(5x) - 4(-4) - x = 0$

Step:4.2.2.3

Complete the distribution.

$5x(5x) + 5x(-4) - 4(5x) - 4(-4) - x = 0$

Step:4.2.3

Combine like terms.

Step:4.2.3.1

Simplify each term.

Step:4.2.3.1.1

Multiply using the commutative property.

$5 \cdot 5x \cdot x + 5x(-4) - 4(5x) - 4(-4) - x = 0$

Step:4.2.3.1.2

Add the exponents when multiplying x by x.

Step:4.2.3.1.2.1

Combine x terms.

$5 \cdot 5(x \cdot x) + 5x(-4) - 4(5x) - 4(-4) - x = 0$

Step:4.2.3.1.2.2

Multiply x by x.

$5 \cdot 5x^{2} + 5x(-4) - 4(5x) - 4(-4) - x = 0$

Step:4.2.3.1.3

Multiply 5 by 5.

$25x^{2} + 5x(-4) - 4(5x) - 4(-4) - x = 0$

Step:4.2.3.1.4

Multiply -4 by 5.

$25x^{2} - 20x - 4(5x) - 4(-4) - x = 0$

Step:4.2.3.1.5

Multiply 5 by -4.

$25x^{2} - 20x - 20x - 4(-4) - x = 0$

Step:4.2.3.1.6

Multiply -4 by -4.

$25x^{2} - 20x - 20x + 16 - x = 0$

Step:4.2.3.2

Combine the x terms.

$25x^{2} - 40x + 16 - x = 0$

Step:4.3

Combine the remaining x terms.

$25x^{2} - 41x + 16 = 0$

Step:5

Factor the quadratic equation.

Step:5.1

Look for two numbers that multiply to $25 \cdot 16 = 400$ and add up to -41.

Step:5.1.1

Factor -41 out of -41x.

$25x^{2} - 41x + 16 = 0$

Step:5.1.2

Split -41 into -16 and -25.

$25x^{2} + (-16 - 25)x + 16 = 0$

Step:5.1.3

Distribute the x term.

$25x^{2} - 16x - 25x + 16 = 0$

Step:5.2

Factor by grouping.

Step:5.2.1

Group the first two and the last two terms.

$(25x^{2} - 16x) - (25x - 16) = 0$

Step:5.2.2

Factor out the GCF from each group.

$x(25x - 16) - 1(25x - 16) = 0$

Step:5.3

Factor out the common binomial factor.

$(25x - 16)(x - 1) = 0$

Step:6

Apply the zero-product property.

$25x - 16 = 0$ or $x - 1 = 0$

Step:7

Solve for x when $25x - 16 = 0$.

Step:7.1

Set the expression equal to zero.

$25x - 16 = 0$

Step:7.2

Isolate x.

Step:7.2.1

Add 16 to both sides.

$25x = 16$

Step:7.2.2

Divide by 25.

Step:7.2.2.1

Divide both sides by 25.

$\frac{25x}{25} = \frac{16}{25}$

Step:7.2.2.2

Simplify the equation.

$x = \frac{16}{25}$

Step:8

Solve for x when $x - 1 = 0$.

Step:8.1

Set the expression equal to zero.

$x - 1 = 0$

Step:8.2

Add 1 to both sides.

$x = 1$

Step:9

Combine the solutions.

$x = \frac{16}{25}$ or $x = 1$

Step:10

Present the solution in various forms.

Exact Form: $x = \frac{16}{25}$ or $x = 1$ Decimal Form: $x = 0.64$ or $x = 1$

Knowledge Notes:

1. Fractional Exponents: The expression $a^{\frac{m}{n}}$ is equivalent to the nth root of $a$ raised to the power m, i.e., $\sqrt[n]{a^m}$.

2. Power of a Power Rule: When raising a power to another power, multiply the exponents, $(a^m)^n = a^{mn}$.

3. Zero-Product Property: If the product of two factors is zero, then at least one of the factors must be zero, i.e., if $ab = 0$, then $a = 0$ or $b = 0$.

4. FOIL Method: A technique for multiplying two binomials, which stands for First, Outer, Inner, Last, representing the terms that are multiplied together.

5. Factoring by Grouping: A method used to factor polynomials with four terms by grouping pairs of terms that have a common factor.

6. Solving Quadratic Equations: Quadratic equations can often be solved by factoring, completing the square, or using the quadratic formula. In this case, factoring was used.