Answer :
To solve the expression \(\frac{49 \times x^{-3}}{7^{-3} \times 10 \times x^{-7}}\), follow these steps:
1. Rewrite the constants in terms of powers of their bases:
[tex]\[ 49 = 7^2 \][/tex]
So, the expression becomes:
[tex]\[ \frac{7^2 \times x^{-3}}{7^{-3} \times 10 \times x^{-7}} \][/tex]
2. Simplify the expression by combining like terms with similar bases:
- For the base 7:
[tex]\[ \frac{7^2}{7^{-3}} = 7^{2 - (-3)} = 7^{2 + 3} = 7^5 \][/tex]
- For the base \(x\):
[tex]\[ \frac{x^{-3}}{x^{-7}} = x^{-3 - (-7)} = x^{-3 + 7} = x^4 \][/tex]
3. Substitute the simplified terms back into the fraction:
[tex]\[ \frac{7^5 \times x^4}{10} \][/tex]
So, the simplified form of the given expression is:
[tex]\[ \frac{7^5 \times x^4}{10} \][/tex]
Therefore, \(\frac{49 \times x^{-3}}{7^{-3} \times 10 \times x^{-7}}\) simplifies to:
[tex]\[ \frac{7^5 x^4}{10} \][/tex]
1. Rewrite the constants in terms of powers of their bases:
[tex]\[ 49 = 7^2 \][/tex]
So, the expression becomes:
[tex]\[ \frac{7^2 \times x^{-3}}{7^{-3} \times 10 \times x^{-7}} \][/tex]
2. Simplify the expression by combining like terms with similar bases:
- For the base 7:
[tex]\[ \frac{7^2}{7^{-3}} = 7^{2 - (-3)} = 7^{2 + 3} = 7^5 \][/tex]
- For the base \(x\):
[tex]\[ \frac{x^{-3}}{x^{-7}} = x^{-3 - (-7)} = x^{-3 + 7} = x^4 \][/tex]
3. Substitute the simplified terms back into the fraction:
[tex]\[ \frac{7^5 \times x^4}{10} \][/tex]
So, the simplified form of the given expression is:
[tex]\[ \frac{7^5 \times x^4}{10} \][/tex]
Therefore, \(\frac{49 \times x^{-3}}{7^{-3} \times 10 \times x^{-7}}\) simplifies to:
[tex]\[ \frac{7^5 x^4}{10} \][/tex]
