To solve the expression [tex]\(\left(49^{-2}\right)^{\frac{1}{3}} \times 7^{-\frac{1}{3}}\)[/tex], follow these steps:
1. Evaluate [tex]\((49^{-2})^{\frac{1}{3}}\)[/tex]:
- First, recall that 49 can be written as [tex]\(7^2\)[/tex].
- So, [tex]\(49^{-2} = (7^2)^{-2}\)[/tex].
- Using the property of exponents [tex]\((a^m)^n = a^{mn}\)[/tex], we have [tex]\((7^2)^{-2} = 7^{2 \cdot -2} = 7^{-4}\)[/tex].
- Now, we need to evaluate [tex]\((7^{-4})^{\frac{1}{3}}\)[/tex].
- Again, applying the exponent rule, [tex]\((a^m)^n = a^{mn}\)[/tex], we get [tex]\(7^{-4 \cdot \frac{1}{3}} = 7^{-\frac{4}{3}}\)[/tex].
2. Evaluate [tex]\(7^{-\frac{1}{3}}\)[/tex]:
- This is already in a simplified form: [tex]\(7^{-\frac{1}{3}}\)[/tex].
3. Combine the two parts:
- Our expression now is [tex]\(7^{-\frac{4}{3}} \times 7^{-\frac{1}{3}}\)[/tex].
- Using the property of exponents [tex]\(a^m \times a^n = a^{m+n}\)[/tex], we combine the exponents:
[tex]\[
7^{-\frac{4}{3}} \times 7^{-\frac{1}{3}} = 7^{-\frac{4}{3} - \frac{1}{3}} = 7^{-\frac{5}{3}}
\][/tex]
4. Simplify the exponent:
- The exponent is [tex]\(-\frac{5}{3}\)[/tex], which means we are finding the reciprocal of the cube root of [tex]\(7^5\)[/tex].
5. Obtain a numerical value:
- Given the pre-computed results, the numerical values are as follows:
[tex]\[
(49^{-2})^{\frac{1}{3}} \approx 0.07468
\][/tex]
and
[tex]\[
7^{-\frac{1}{3}} \approx 0.52276
\][/tex]
- Multiplying these results together, we get:
[tex]\[
0.07468 \times 0.52276 \approx 0.03904
\][/tex]
Therefore, the evaluated expression [tex]\(\left(49^{-2}\right)^{\frac{1}{3}} \times 7^{-\frac{1}{3}}\)[/tex] results in approximately [tex]\(0.03904\)[/tex].
