Simplify radicals and exponents $\sqrt[3]{({x}^{3}\text{}y{)}^{2}{y}^{4}}$

zi2lalZ
2021-03-01
Answered

Simplify radicals and exponents $\sqrt[3]{({x}^{3}\text{}y{)}^{2}{y}^{4}}$

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Ayesha Gomez

Answered 2021-03-02
Author has **104** answers

Distribute the exponent 2 inside the bracket.

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

$\sqrt[3]{({x}^{3}\text{}y{)}^{2}{y}^{4}}$

$=\sqrt[3]{{x}^{3\text{}\times \text{}2}{y}^{2}{y}^{4}}$

$=\sqrt[3]{{x}^{6}{y}^{2}{y}^{4}}$

Since the bases are for y, so add the exponents.

Formula:${a}^{b}\text{}\cdot \text{}{a}^{c}={a}^{b+c}$

$=\sqrt[3]{{x}^{6}{y}^{2}{y}^{4}}$

$=\sqrt[3]{{x}^{6}{y}^{2\text{}+\text{}4}}$

$=\sqrt[3]{{x}^{6}{y}^{6}}$

Because of the cube root we have to divide the exponents by 3.

$\sqrt[3]{{x}^{6}{y}^{6}}$

$={x}^{\frac{6}{3}}{y}^{\frac{6}{3}}$

$={x}^{2}{y}^{2}$

Finally answer:${x}^{2}{y}^{2}$

Formula:

Since the bases are for y, so add the exponents.

Formula:

Because of the cube root we have to divide the exponents by 3.

Finally answer:

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I have equations for two lines, one of which is linear and the other is logarithmic, ie:

..and I need to find out where (if at all) these lines intersect. I realise that I need to solve:

..for x, but apart from shuffling the constants around I'm not sure how to do this

Is there a general solution to this problem?

Thanks

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