What is the answer to cube root of 2x-1}=\sqrt[6]{x+1 ?

The answer to cube root of 2x-1}=\sqrt[6]{x+1 is x= 5/4

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cube root of 2x-1}=\sqrt[6]{x+1

x = \frac{5}{4}

Decimal Notation

x = 1.25

Solution steps

Take both sides of the equation to the power of

6 : (2 x - 1)^{2} = x + 1

(2 x - 1)^{2} = x + 1

Solve

(2 x - 1)^{2} = x + 1 : x = \frac{5}{4}, x = 0

x = \frac{5}{4}, x = 0

Verify Solutions:

x = \frac{5}{4}

True

, x = 0

False

The solution is

x = \frac{5}{4}

20 - x = - x

G (x) = x^{2} - 81

\frac{x}{10} = 1

y, - y + 30 = y

y + 3 = 2 y - 3

What is the answer to cube root of 2x-1}=\sqrt[6]{x+1 ?
The answer to cube root of 2x-1}=\sqrt[6]{x+1 is x= 5/4
Solve for x: cube root of 2x-1}=\sqrt[6]{x+1
The solution is x= 5/4