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simplify the expression. $\frac{b^{-\frac{4}{3}} b^{\frac{1}{2}}}{b^{\f…

Question

simplify the expression.
$\frac{b^{-\frac{4}{3}} b^{\frac{1}{2}}}{b^{\frac{1}{3}}}$
write your answer using only positive exponents.
assume that all variables are positive real numbers.

Explanation:

Step1: Add exponents in numerator

When multiplying terms with the same base, add exponents: $b^{-\frac{4}{3}}b^{\frac{1}{2}} = b^{-\frac{4}{3}+\frac{1}{2}}$
Calculate the sum: $-\frac{4}{3}+\frac{1}{2} = -\frac{8}{6}+\frac{3}{6}=-\frac{5}{6}$
So numerator becomes $b^{-\frac{5}{6}}$

Step2: Subtract exponents for division

When dividing terms with the same base, subtract denominator exponent from numerator exponent: $\frac{b^{-\frac{5}{6}}}{b^{\frac{1}{3}}}=b^{-\frac{5}{6}-\frac{1}{3}}$
Calculate the difference: $-\frac{5}{6}-\frac{2}{6}=-\frac{7}{6}$
So we get $b^{-\frac{7}{6}}$

Step3: Convert to positive exponent

Use the rule $b^{-n}=\frac{1}{b^n}$: $b^{-\frac{7}{6}}=\frac{1}{b^{\frac{7}{6}}}$

Answer:

$\frac{1}{b^{\frac{7}{6}}}$