Solve: 10^8 + 10^-4 + (1/10)^-16 Expression with Mixed Powers

Let's use the law of exponents for negative exponents:

$a^{-n} = \frac{1}{a^n}$and apply this law to the problem:

$10^8+10^{-4}+(\frac{1}{10})^{-16}=10^8+\frac{1}{10^4}+(10^{-1})^{-16}$$$when we apply the above law of exponents to the second term in the sum, and the same law but in the opposite direction - we'll apply it to the fraction inside the parentheses of the third term in the sum,

Now let's recall the law of exponents for exponent of an exponent:

$(a^m)^n=a^{m\cdot n}$we'll apply this law to the expression we got in the last step:

$10^8+\frac{1}{10^4}+(10^{-1})^{-16}=10^8+\frac{1}{10^4}+10^{(-1)\cdot(-16)}=10^8+\frac{1}{10^4}+10^{16}$when we apply this law to the third term from the left and then simplify the resulting expression,

Let's summarize the solution steps, we got that:

$10^8+10^{-4}+(\frac{1}{10})^{-16}=10^8+\frac{1}{10^4}+(10^{-1})^{-16} =10^8+\frac{1}{10^4}+10^{16}$Therefore the correct answer is answer A.