= Solution
With $R_f=R_i$, the output and input voltage magnitudes are equal. The signal power supplied at the input is $P_i=V_{i,\rm rms}^2/R_i$, while the load receives $P_L=V_{o,\rm rms}^2/R_L$. Hence
$$
\boxed{\frac{P_L}{P_i}=\left(\frac{R_f}{R_i}\right)^2\frac{R_i}{R_L}=\frac{1000}{100}=10.}
$$
This is $10$ dB of power gain into the specified load. Output energy comes from the amplifier's supply rails; unity voltage-gain magnitude does not imply unity power gain. Dissipation in the feedback resistor is additional to the specified load power.
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