Math / Algebra

Question $(7 x-5)^{\frac{2}{3}}-5(7 x-5)^{\frac{-1}{3}}=0$

Studdy Solution

STEP 1

1. The expression $(7x-5)^{\frac{2}{3}}-5(7x-5)^{\frac{-1}{3}}=0$ is an equation involving fractional exponents.
2. We can solve this equation by making a substitution to simplify the exponents.
3. The goal is to find the value(s) of $x$ that satisfy the equation.

STEP 2

1. Make a substitution to transform the equation into a simpler form.
2. Solve the resulting equation for the new variable.
3. Substitute back to find the value(s) of $x$ .

STEP 3

Make the substitution $u = (7x-5)^{\frac{1}{3}}$ .
Then the equation becomes: $u^2 - 5u^{-1} = 0$

STEP 4

Multiply both sides of the equation by $u$ to eliminate the negative exponent: $u^3 - 5 = 0$

STEP 5

Solve for $u$ by isolating $u^3$ : $u^3 = 5$

STEP 6

Take the cube root of both sides to solve for $u$ : $u = \sqrt[3]{5}$

STEP 7

Substitute back $u = (7x-5)^{\frac{1}{3}}$ to return to the original variable $x$ : $(7x-5)^{\frac{1}{3}} = \sqrt[3]{5}$

STEP 8

Raise both sides to the power of 3 to solve for $7x-5$ : $7x - 5 = (\sqrt[3]{5})^3$ $7x - 5 = 5$

STEP 9

Solve for $x$ : $7x = 10$ $x = \frac{10}{7}$
Solution: The value of $x$ that satisfies the equation is: $x = \frac{10}{7}$

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Question $(7 x-5)^{\frac{2}{3}}-5(7 x-5)^{\frac{-1}{3}}=0$

Studdy Solution

STEP 1

1. The expression $(7x-5)^{\frac{2}{3}}-5(7x-5)^{\frac{-1}{3}}=0$ is an equation involving fractional exponents.
2. We can solve this equation by making a substitution to simplify the exponents.
3. The goal is to find the value(s) of $x$ that satisfy the equation.

STEP 2

1. Make a substitution to transform the equation into a simpler form.
2. Solve the resulting equation for the new variable.
3. Substitute back to find the value(s) of $x$ .

STEP 3

Make the substitution $u = (7x-5)^{\frac{1}{3}}$ .
Then the equation becomes: $u^2 - 5u^{-1} = 0$

STEP 4

Multiply both sides of the equation by $u$ to eliminate the negative exponent: $u^3 - 5 = 0$

STEP 5

Solve for $u$ by isolating $u^3$ : $u^3 = 5$

STEP 6

Take the cube root of both sides to solve for $u$ : $u = \sqrt[3]{5}$

STEP 7

Substitute back $u = (7x-5)^{\frac{1}{3}}$ to return to the original variable $x$ : $(7x-5)^{\frac{1}{3}} = \sqrt[3]{5}$

STEP 8

Raise both sides to the power of 3 to solve for $7x-5$ : $7x - 5 = (\sqrt[3]{5})^3$ $7x - 5 = 5$

STEP 9

Solve for $x$ : $7x = 10$ $x = \frac{10}{7}$
Solution: The value of $x$ that satisfies the equation is: $x = \frac{10}{7}$

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Question(7x−5)23−5(7x−5)−13=0(7 x-5)^{\frac{2}{3}}-5(7 x-5)^{\frac{-1}{3}}=0(7x−5)32​−5(7x−5)3−1​=0

Studdy Solution

STEP 1

STEP 2

1. Make a substitution to transform the equation into a simpler form.2. Solve the resulting equation for the new variable.3. Substitute back to find the value(s) of xxx.

STEP 3

Make the substitution u=(7x−5)13u = (7x-5)^{\frac{1}{3}}u=(7x−5)31​.Then the equation becomes: u2−5u−1=0 u^2 - 5u^{-1} = 0 u2−5u−1=0

STEP 4

Multiply both sides of the equation by uuu to eliminate the negative exponent: u3−5=0 u^3 - 5 = 0 u3−5=0

STEP 5

Solve for uuu by isolating u3u^3u3: u3=5 u^3 = 5 u3=5

STEP 6

Take the cube root of both sides to solve for uuu: u=53 u = \sqrt[3]{5} u=35​

STEP 7

Substitute back u=(7x−5)13u = (7x-5)^{\frac{1}{3}}u=(7x−5)31​ to return to the original variable xxx: (7x−5)13=53 (7x-5)^{\frac{1}{3}} = \sqrt[3]{5} (7x−5)31​=35​

STEP 8

Raise both sides to the power of 3 to solve for 7x−57x-57x−5: 7x−5=(53)3 7x - 5 = (\sqrt[3]{5})^3 7x−5=(35​)3 7x−5=5 7x - 5 = 5 7x−5=5

STEP 9

Solve for xxx: 7x=10 7x = 10 7x=10 x=107 x = \frac{10}{7} x=710​Solution: The value of xxx that satisfies the equation is: x=107 x = \frac{10}{7} x=710​

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Question $(7 x-5)^{\frac{2}{3}}-5(7 x-5)^{\frac{-1}{3}}=0$

1. Make a substitution to transform the equation into a simpler form.
2. Solve the resulting equation for the new variable.
3. Substitute back to find the value(s) of $x$ .

Make the substitution $u = (7x-5)^{\frac{1}{3}}$ .
Then the equation becomes: $u^2 - 5u^{-1} = 0$

Multiply both sides of the equation by $u$ to eliminate the negative exponent: $u^3 - 5 = 0$

Solve for $u$ by isolating $u^3$ : $u^3 = 5$

Take the cube root of both sides to solve for $u$ : $u = \sqrt[3]{5}$

Substitute back $u = (7x-5)^{\frac{1}{3}}$ to return to the original variable $x$ : $(7x-5)^{\frac{1}{3}} = \sqrt[3]{5}$

Raise both sides to the power of 3 to solve for $7x-5$ : $7x - 5 = (\sqrt[3]{5})^3$ $7x - 5 = 5$

Solve for $x$ : $7x = 10$ $x = \frac{10}{7}$
Solution: The value of $x$ that satisfies the equation is: $x = \frac{10}{7}$