SL 1.5—Exponents and Logarithms

The expression involves multiplying and dividing powers with the same base (7), so we apply the Product Rule and Quotient Rule.

$7^5\times 7^{-3}=7^{5+(-3)}=7^2$

$\frac{7^2}{7^{-1}}=7^{2-(-1)}=7^{2+1}=7^3$

The simplified expression is $7^3$. Note that subtracting a negative exponent adds it back, which is a common source of error.

We need to simplify $\frac{7^5\times 7^{-3}}{7^{-1}}$ to a single power of 7. We can add all exponents in the numerator and subtract the denominator exponent.

This would result from computing $5+(-3)-(-1)=5-3-1=1$, which incorrectly subtracts the negative exponent instead of adding it. Eliminated.

This would come from $5-3-(-1)-2=-1$, which does not follow from any correct application of the laws. Eliminated.

$7^9$ would result from $5+3+1=9$, incorrectly treating all exponents as positive. Eliminated.

The correct exponent is $5+(-3)-(-1)=5-3+1=3$, giving $7^3$.

A negative exponent does not make the result negative. The rule states: $x^{-a}=\frac{1}{x^a}$

$4^{-3}=\frac{1}{4^3}$

$4^3=4\times 4\times 4=64$

Therefore $4^{-3}=\frac{1}{64}$.

We must evaluate $4^{-3}$, a negative exponent. A negative exponent means reciprocal — it does not produce a negative number.

$-64$ would incorrectly interpret a negative exponent as making the result negative. This is a common misconception. Eliminated.

$\frac{1}{12}$ would come from $\frac{1}{4\times 3}=\frac{1}{12}$, incorrectly multiplying base by exponent instead of raising to the power. Eliminated.

$64=4^3$ ignores the negative sign on the exponent entirely. Eliminated.

$4^{-3}=\frac{1}{4^3}=\frac{1}{64}$ by the negative exponent rule.