Solve for t 6t^(2+5)=2t^(2+1)

$6 t^{2 + 5} = 2 t^{2 + 1}$

Add $2$ and $5$ .

$6 t^{7} = 2 t^{2 + 1}$

Add $2$ and $1$ .

$6 t^{7} = 2 t^{3}$

Subtract $2 t^{3}$ from both sides of the equation.

$6 t^{7} - 2 t^{3} = 0$

Factor $2 t^{3}$ out of $6 t^{7} - 2 t^{3}$ .

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Factor $2 t^{3}$ out of $6 t^{7}$ .

$2 t^{3} (3 t^{4}) - 2 t^{3} = 0$

Factor $2 t^{3}$ out of $- 2 t^{3}$ .

$2 t^{3} (3 t^{4}) + 2 t^{3} (- 1) = 0$

Factor $2 t^{3}$ out of $2 t^{3} (3 t^{4}) + 2 t^{3} (- 1)$ .

$2 t^{3} (3 t^{4} - 1) = 0$

Divide each term by $2$ and simplify.

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Divide each term in $2 t^{3} (3 t^{4} - 1) = 0$ by $2$ .

$\frac{2 t^{3} (3 t^{4} - 1)}{2} = \frac{0}{2}$

Simplify $\frac{2 t^{3} (3 t^{4} - 1)}{2}$ .

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Simplify terms.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 t^{3} (3 t^{4} - 1)}{2} = \frac{0}{2}$

Divide $t^{3} (3 t^{4} - 1)$ by $1$ .

$t^{3} (3 t^{4} - 1) = \frac{0}{2}$

Apply the distributive property.

$t^{3} (3 t^{4}) + t^{3} \cdot - 1 = \frac{0}{2}$

Reorder.

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Rewrite using the commutative property of multiplication.

$3 t^{3} t^{4} + t^{3} \cdot - 1 = \frac{0}{2}$

Move $- 1$ to the left of $t^{3}$ .

$3 t^{3} t^{4} - 1 \cdot t^{3} = \frac{0}{2}$

Simplify each term.

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Multiply $t^{3}$ by $t^{4}$ by adding the exponents.

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Move $t^{4}$ .

$3 (t^{4} t^{3}) - 1 \cdot t^{3} = \frac{0}{2}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3 t^{4 + 3} - 1 \cdot t^{3} = \frac{0}{2}$

Add $4$ and $3$ .

$3 t^{7} - 1 \cdot t^{3} = \frac{0}{2}$

Rewrite $- 1 t^{3}$ as $- t^{3}$ .

$3 t^{7} - t^{3} = \frac{0}{2}$

Divide $0$ by $2$ .

$3 t^{7} - t^{3} = 0$

Factor $t^{3}$ out of $3 t^{7} - t^{3}$ .

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Factor $t^{3}$ out of $3 t^{7}$ .

$t^{3} (3 t^{4}) - t^{3} = 0$

Factor $t^{3}$ out of $- t^{3}$ .

$t^{3} (3 t^{4}) + t^{3} \cdot - 1 = 0$

Factor $t^{3}$ out of $t^{3} (3 t^{4}) + t^{3} \cdot - 1$ .

$t^{3} (3 t^{4} - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$t^{3} = 0$

$3 t^{4} - 1 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$t^{3} = 0$

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$t = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

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Rewrite $0$ as $0^{3}$ .

$t = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$t = 0$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$3 t^{4} - 1 = 0$

Add $1$ to both sides of the equation.

$3 t^{4} = 1$

Divide each term by $3$ and simplify.

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Divide each term in $3 t^{4} = 1$ by $3$ .

$\frac{3 t^{4}}{3} = \frac{1}{3}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 t^{4}}{3} = \frac{1}{3}$

Divide $t^{4}$ by $1$ .

$t^{4} = \frac{1}{3}$

Take the 4th root of both sides of the equation to eliminate the exponent on the left side.

$t = \pm \sqrt[4]{\frac{1}{3}}$

The complete solution is the result of both the positive and negative portions of the solution.

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Simplify the right side of the equation.

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Rewrite $\sqrt[4]{\frac{1}{3}}$ as $\frac{\sqrt[4]{1}}{\sqrt[4]{3}}$ .

$t = \pm \frac{\sqrt[4]{1}}{\sqrt[4]{3}}$

Any root of $1$ is $1$ .

$t = \pm \frac{1}{\sqrt[4]{3}}$

Multiply $\frac{1}{\sqrt[4]{3}}$ by $\frac{{\sqrt[4]{3}}^{3}}{{\sqrt[4]{3}}^{3}}$ .

$t = \pm \frac{1}{\sqrt[4]{3}} \cdot \frac{{\sqrt[4]{3}}^{3}}{{\sqrt[4]{3}}^{3}}$

Combine and simplify the denominator.

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Multiply $\frac{1}{\sqrt[4]{3}}$ and $\frac{{\sqrt[4]{3}}^{3}}{{\sqrt[4]{3}}^{3}}$ .

$t = \pm \frac{{\sqrt[4]{3}}^{3}}{\sqrt[4]{3} {\sqrt[4]{3}}^{3}}$

Raise $\sqrt[4]{3}$ to the power of $1$ .

$t = \pm \frac{{\sqrt[4]{3}}^{3}}{\sqrt[4]{3} {\sqrt[4]{3}}^{3}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$t = \pm \frac{{\sqrt[4]{3}}^{3}}{{\sqrt[4]{3}}^{1 + 3}}$

Add $1$ and $3$ .

$t = \pm \frac{{\sqrt[4]{3}}^{3}}{{\sqrt[4]{3}}^{4}}$

Rewrite ${\sqrt[4]{3}}^{4}$ as $3$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{3}$ as $3^{\frac{1}{4}}$ .

$t = \pm \frac{{\sqrt[4]{3}}^{3}}{{(3^{\frac{1}{4}})}^{4}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$t = \pm \frac{{\sqrt[4]{3}}^{3}}{3^{\frac{1}{4} \cdot 4}}$

Combine $\frac{1}{4}$ and $4$ .

$t = \pm \frac{{\sqrt[4]{3}}^{3}}{3^{\frac{4}{4}}}$

Cancel the common factor of $4$ .

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Cancel the common factor.

$t = \pm \frac{{\sqrt[4]{3}}^{3}}{3^{\frac{4}{4}}}$

Divide $1$ by $1$ .

$t = \pm \frac{{\sqrt[4]{3}}^{3}}{3}$

Evaluate the exponent.

$t = \pm \frac{{\sqrt[4]{3}}^{3}}{3}$

Simplify the numerator.

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Rewrite ${\sqrt[4]{3}}^{3}$ as ${(3^{3})}^{\frac{1}{4}}$ .

$t = \pm \frac{\sqrt[4]{3^{3}}}{3}$

Raise $3$ to the power of $3$ .

$t = \pm \frac{\sqrt[4]{27}}{3}$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$t = \frac{\sqrt[4]{27}}{3}$

Next, use the negative value of the $\pm$ to find the second solution.

$t = - \frac{\sqrt[4]{27}}{3}$

The complete solution is the result of both the positive and negative portions of the solution.

$t = \frac{\sqrt[4]{27}}{3}, - \frac{\sqrt[4]{27}}{3}$

The final solution is all the values that make $t^{3} (3 t^{4} - 1) = 0$ true.

$t = 0, \frac{\sqrt[4]{27}}{3}, - \frac{\sqrt[4]{27}}{3}$

The result can be shown in multiple forms.

Exact Form:

$t = 0, \frac{\sqrt[4]{27}}{3}, - \frac{\sqrt[4]{27}}{3}$

Decimal Form:

$t = 0, 0.75983568 \dots, - 0.75983568 \dots$

Solve for t 6t^(2+5)=2t^(2+1)

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