How do you solve ${{3 x - 2)}^{\frac{3}{5}} = 8$ ${3x-2)^(3/5) = 8$ ?

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How do you solve ${{3 x - 2)}^{\frac{3}{5}} = 8$ ${3x-2)^(3/5) = 8$ ?

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Answer 1 · 2016-04-26T18:11:02

$x = \frac{34}{3}$ $x=34/3$

Explanation:

Our first step should be to undo the $\frac{3}{5}$ $3/5$ power. In order to do this, raise each side to the $\frac{5}{3}$ $5/3$ power.

${({(3 x - 2)}^{\frac{3}{5}})}^{\frac{5}{3}} = 8^{\frac{5}{3}}$ $((3x-2)^(3/5))^(5/3)=8^(5/3)$

On the left hand side, use the following rule:

${(a^{b})}^{c} = a^{b c}$ $(a^b)^c=a^(bc)$

Applying this rule to the left side, we multiply $\frac{3}{5}$ $3/5$ and $\frac{5}{3}$ $5/3$ :

${(3 x - 2)}^{\frac{3}{5} \times \frac{5}{3}} = 8^{\frac{5}{3}}$ $(3x-2)^(3/5xx5/3)=8^(5/3)$

${(3 x - 2)}^{1} = 8^{\frac{5}{3}}$ $(3x-2)^1=8^(5/3)$

$3 x - 2 = 8^{\frac{5}{3}}$ $3x-2=8^(5/3)$

Now, we should simplify $8^{\frac{5}{3}}$ $8^(5/3)$ . Do this by writing $8$ $8$ as $2^{3}$ $2^3$ .

$3 x - 2 = {(2^{3})}^{\frac{5}{3}}$ $3x-2=(2^3)^(5/3)$

Use the exponent rule again.

$3 x - 2 = 2^{(3 \times \frac{5}{3})}$ $3x-2=2^((3xx5/3))$

$3 x - 2 = 2^{5}$ $3x-2=2^5$

$3 x - 2 = 32$ $3x-2=32$

$3 x = 34$ $3x=34$

$x = \frac{34}{3}$ $x=34/3$

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1 Answer

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