Solve sqrt(( 7/1 )^(x 1)-( 1/7 )^(2x^3))

Evaluate

(\frac{7}{1})^{x \times 1} - (\frac{1}{7})^{2 x^{3}}

To find the roots of the expression,set the expression equal to 0

(\frac{7}{1})^{x \times 1} - (\frac{1}{7})^{2 x^{3}} = 0

Find the domain

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Evaluate

(\frac{7}{1})^{x \times 1} - (\frac{1}{7})^{2 x^{3}} \geq 0

Simplify

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7^{x} - 7^{- 2 x^{3}} \geq 0

Factor the expression

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(7^{x + 2 x^{3}} - 1) (7^{(x^{3})})^{- 2} \geq 0

Expand the expression

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\frac{7 ^{x + 2 x^{3}} - 1}{( 7 ^{(x^{3})} ) ^{2}} \geq 0

Separate the inequality into 2 possible cases

⎩ ⎨ ⎧ 7^{x + 2 x^{3}} - 1 \geq 0 (7^{(x^{3})})^{2} > 0 ⎩ ⎨ ⎧ 7^{x + 2 x^{3}} - 1 \leq 0 (7^{(x^{3})})^{2} < 0

Solve the inequality

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{x \geq 0 (7^{(x^{3})})^{2} > 0 ⎩ ⎨ ⎧ 7^{x + 2 x^{3}} - 1 \leq 0 (7^{(x^{3})})^{2} < 0

Since the left-hand side is always positive,and the right-hand side is always 0,the statement is true for any value of x

{x \geq 0 x \in R ⎩ ⎨ ⎧ 7^{x + 2 x^{3}} - 1 \leq 0 (7^{(x^{3})})^{2} < 0

Solve the inequality

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{x \geq 0 x \in R {x \leq 0 (7^{(x^{3})})^{2} < 0

Since the left-hand side is always positive,and the right-hand side is always 0,the statement is false for any value of x

{x \geq 0 x \in R {x \leq 0 x \in / R

Find the intersection

x \geq 0 {x \leq 0 x \in / R

Find the intersection

x \geq 0 x \in / R

Find the union

x \geq 0

(\frac{7}{1})^{x \times 1} - (\frac{1}{7})^{2 x^{3}} = 0, x \geq 0

Calculate

(\frac{7}{1})^{x \times 1} - (\frac{1}{7})^{2 x^{3}} = 0

Divide the terms

7^{x \times 1} - (\frac{1}{7})^{2 x^{3}} = 0

Any expression multiplied by 1 remains the same

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

Rewrite the expression

7^{x} - 7^{- 2 x^{3}} = 0

Rewrite the expression

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\frac{7 ^{x + 2 x^{3}} - 1}{7 ^{2 x^{3}}} = 0

The only way a root could be 0 is when the radicand equals 0

\frac{7 ^{x + 2 x^{3}} - 1}{7 ^{2 x^{3}}} = 0

Cross multiply

7^{x + 2 x^{3}} - 1 = 7^{2 x^{3}} \times 0

Simplify the equation

7^{x + 2 x^{3}} - 1 = 0

Move the expression to the right side

7^{x + 2 x^{3}} = 1

Rewrite in exponential form

7^{x + 2 x^{3}} = 7^{0}

Since the bases are the same,set the exponents equal

x + 2 x^{3} = 0

Factor the expression

x (1 + 2 x^{2}) = 0

Separate the equation into 2 possible cases

x = 0 1 + 2 x^{2} = 0

Solve the equation

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x = 0 x \in / R

Find the union

x = 0

Check if the solution is in the defined range

x = 0, x \geq 0

Solution

x = 0

Simplify the expression