Question

simplify.
9^{\frac{1}{4}}cdot9^{\frac{1}{3}}

Explanation:

Step1: Use exponent - product rule

According to the rule \(a^m\cdot a^n=a^{m + n}\), for \(a = 9\), \(m=\frac{1}{4}\), and \(n=\frac{1}{3}\), we have \(9^{\frac{1}{4}}\cdot9^{\frac{1}{3}}=9^{\frac{1}{4}+\frac{1}{3}}\).

Step2: Calculate the sum of the exponents

Find a common denominator for \(\frac{1}{4}+\frac{1}{3}\). The common denominator of 4 and 3 is 12. So \(\frac{1}{4}+\frac{1}{3}=\frac{1\times3}{4\times3}+\frac{1\times4}{3\times4}=\frac{3 + 4}{12}=\frac{7}{12}\). Then \(9^{\frac{1}{4}+\frac{1}{3}}=9^{\frac{7}{12}}\). Since \(9 = 3^2\), we can rewrite it as \((3^2)^{\frac{7}{12}}\).

Step3: Use the power - of - a - power rule

According to the rule \((a^m)^n=a^{mn}\), for \(a = 3\), \(m = 2\), and \(n=\frac{7}{12}\), we have \((3^2)^{\frac{7}{12}}=3^{2\times\frac{7}{12}}=3^{\frac{7}{6}}\).

Answer:

\(3^{\frac{7}{6}}\)