Simplify 4a^-10 times a times d, writing the answer with positive exponents.

Step-by-step solution

1. 1

   Spot the hidden exponent. The lone $a$ is really $a^{1}$ — an unwritten exponent of $1$. Making it explicit is what allows the exponent rule to apply:

   $4a^{-10}\cdot a^{1}\cdot d$

   The factors $4$ and $d$ carry no $a$, so they are simply spectators in this step.

2. 2

   Add the exponents of the shared base. The product rule $a^{m}\cdot a^{n}=a^{m+n}$ applies only to powers of the same base:

   $a^{-10}\cdot a^{1}=a^{-10+1}=a^{-9}$

   Exponents are added, not multiplied — $a^{-10}\cdot a^{1}$ is not $a^{-10}$. The expression is now $4a^{-9}d$.

3. 3

   Convert the negative exponent. By definition $a^{-n}=\dfrac{1}{a^{n}}$, so a negative exponent signals that the factor belongs on the other side of the fraction bar:

   $4a^{-9}d=\frac{4d}{a^{9}}$

   Only $a$ moves. The $4$ and the $d$ have positive exponents already and stay in the numerator — a frequent error is dragging the coefficient $4$ down with it.

4. 4

   State the domain restriction. The original expression contains $a^{-10}$, which is undefined at $a=0$, and so is $\dfrac{4d}{a^{9}}$ — consistently. The simplification is valid for every $a\neq0$ and every $d$.

5. 5

   Verify numerically. Comparing $4a^{-10}\cdot a\cdot d$ with $\dfrac{4d}{a^{9}}$ at $a=1.3,\,2.7,\,-1.9$ and $d=2,\,-3.5$ gives agreement to within $10^{-9}$ in all six combinations ✓. Note that odd exponent $9$ preserves the sign of $a$, so the result is negative when $a<0$ and $d>0$.