How do you simplify $5^{\frac{2}{3}} \cdot 5^{\frac{4}{3}}$ $5^(2/3) * 5^(4/3)$ ?

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How do you simplify $5^{\frac{2}{3}} \cdot 5^{\frac{4}{3}}$ $5^(2/3) * 5^(4/3)$ ?

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Answer 1 · 2018-04-05T02:27:11

$5^{\frac{2}{3}} \cdot 5^{\frac{4}{3}}$ $5^(2/3) * 5^(4/3)$ can be simplified to $25$ $25$ .

Explanation:

By exponent laws, $x^{a} \cdot x^{b} = x^{a + b}$ $x^a*x^b = x^(a+b)$ .

$5^{\frac{2}{3}} \cdot 5^{\frac{4}{3}}$ $5^(2/3) * 5^(4/3)$
$= 5^{\frac{2}{3} + \frac{4}{3}}$ $=5^(2/3+4/3)$
$= 5^{\frac{6}{3}}$ $=5^(6/3)$
$= 5^{2}$ $=5^2$
$= 25$ $=25$

Answer 2 · 2018-04-05T02:27:58

$\Rightarrow 5^{2}$ $color(green)(=> 5^(2)$

Explanation:

![https://www.youtube.com/watch?v=ARLS2TmFT94]( useruploads.socratic.org )

Given $5^{\frac{2}{3}} \cdot 5^{\frac{4}{3}}$ $5^(2/3) * 5 ^ (4/3)$

As per the theory of inices,

$x^{m} \cdot x^{n} = x^{m + n}$ $x^m * x ^n = x^ (m + n)$

$:. => 5 ^ ((2/3) + (4/3))$

$=> 5 ^ ((2+4)/3), " taking 3 as LCM "$

$=> 5^(6/3) " or " 5^(2)$

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How do you simplify $5^{\frac{2}{3}} \cdot 5^{\frac{4}{3}}$ $5^(2/3) * 5^(4/3)$ ?

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How do you simplify $5^{\frac{2}{3}} \cdot 5^{\frac{4}{3}}$ $5^(2/3) * 5^(4/3)$ ?