Question

solve
$\sqrt[4]{2x - 1} = \sqrt[3]{8}$
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Answer 1

ur question is solved above

Answer 2

Given equation,



$\sqrt[4]{2x - 1} = \sqrt[3]{8}$

$\sqrt[4]{2x - 1} = 2$

$2x - 1 = 2^4$

2x - 1 = 16

2x = 16 + 1

2x = 17

x = $\frac{17}{2}$

$\boxed{\green{\bold{x = 8.5}}}$

By substituting the value of x in the given equation.

$\sqrt[4]{2x - 1} = \sqrt[3]{8}$

$\sqrt[4]{2(8.5 - 1)} = \sqrt[3]{8}$

$\sqrt{4}{17 - 1} = \sqrt[3] {8}$

$\sqrt[4]{16} = \sqrt[3]{8}$

2 = 2

$\boxed{\bold{\pink{L.H.S = R.H.S}}}$

Explaination:

Here in the given equation $\sqrt[4]{2x - 1} = \sqrt[3]{8}$, we have to find the value of x.

So, in second step, I'd written the value of $\sqrt[3]{8}$ which is 2.

In step three, I'd taken the 4th root to R.H.S in the form of power raised to 4.

In step four, I'd just written the value of 2⁴ which is 16.

In step five, I'd just taken - 1 to R.H.S which will turn in the form of + 1.

In step six, I'd added 16 and 1.

In step seven, I'd taken 2 to R.H.S.

In step eight, I'd divided 17 and 2 which is equal to 1.

Then, I'd just substituted the value of x in the given expression $\sqrt[4]{2x - 1} = \sqrt[3]{8}$ to prove the value of x which we gotta is correct or not.

Thus, we get the value of x as 8.5.