Question

# To calculate: To evaluate the expression x^{^5/_3} cdot x^{^{-3}/4}

To calculate:
To evaluate the expression
$$\displaystyle{x}^{{^{5}\angle{3}}}\ \cdot\ {x}^{{^\frac{{-{3}}}{{4}}}}$$

2021-01-23
Calculation:
According to the basic rules of exponents, $$\displaystyle{a}^{{{m}}}\ \cdot\ {a}^{{{n}}}={a}^{{{m}\ +\ {n}}}$$
Applying the above rule,
This can be written as $$\displaystyle{x}^{{^{5}\angle{3}}}\ \cdot\ {x}^{{^\frac{{-{3}}}{{4}}}}={x}^{{{\left\lbrace^{5}\angle{3}\right\rbrace}\ +\ {\left(-{3}\angle{4}\right)}}}$$
The exponent has rational numbers whose denominators are different. Hence the LCM of denominators is to be taken
The LCM of 3 and 4 is 12.
$$\displaystyle{x}^{{^{5}\angle{3}}}\ \cdot\ {x}^{{^\frac{{-{3}}}{{4}}}}={x}^{{{20}\ +\ {\left\lbrace{\left(-{9}\right)}\angle{\left\lbrace{12}\right\rbrace}\right\rbrace}}}$$
$$\displaystyle{x}^{{^{5}\angle{3}}}\ \cdot\ {x}^{{^\frac{{-{3}}}{{4}}}}={x}^{{{20}\ -\ {\left\lbrace{9}\angle{\left\lbrace{12}\right\rbrace}\right\rbrace}}}$$
$$\displaystyle{x}^{{^{5}\angle{3}}}\ \cdot\ {x}^{{^\frac{{-{3}}}{{4}}}}={x}^{{^{\left\lbrace{11}\right\rbrace}\angle{\left\lbrace{12}\right\rbrace}}}$$