Ballparking: Overview

Which of the following is closest to $$\left(0.5\cdot10^5\right)^{\frac{3}{2}}~$$?

Incorrect. [[snippet]] You may have gotten 75,000 if you multiplied by $$\frac{3}{2}$$ instead of using it as an exponent.

Incorrect. [[snippet]] It's possible to get 175,000 by multiplying 0.5 by 10 before applying the exponent of 5. Remember, exponents apply before multiplication, so the exponent of 5 doesn't apply to the 0.5.

Incorrect. [[snippet]] You could get 1,100,000 if you didn't apply the exponent to the power of 10.

Correct! A fractional exponent is another way of writing a root where the numerator of the fraction is the power and the denominator of the fraction is the order of the root. Converting the expression and applying the powers of 10 to the 0.5 will make the answer easier to find. $$\hspace{0.5in} \sqrt{(0.5\cdot10^5)^3} = \sqrt{(0.5\cdot10\cdot10^4)^3} = \sqrt{(5\cdot10^4)^3}$$ Now apply the exponent of 3 using the rules of exponents. $$\hspace{0.5in} \sqrt{(5\cdot10^4)^3} = \sqrt{(5)^3\cdot (10^4)^3} = \sqrt{125\cdot 10^{12}}$$ You can split this into two square roots. $$\hspace{0.5in} \sqrt{125\cdot 10^{12}} = \sqrt{125}\cdot\sqrt{10^{12}} = \sqrt{125}\cdot 10^6$$ Finally,, **Ballpark** $$\sqrt{125}$$ as being between 11 and 12, since $$\hspace{0.5in} 11^2 = 121 \mbox{ and } 12^2=144$$. Therefore, $$\hspace{0.5in} \sqrt{125}\cdot 10^6 \approx 11\cdot10^6 = 11{,}000{,}000$$.

Incorrect. [[snippet]] An answer of 16,000,000 is close to $$\hspace{0.5in} 0.5\cdot\left(10^5\right)^{\frac{3}{2}}$$. But the exponent of $$\frac{3}{2}$$ in this problem applies to the 0.5 as well as the power of 10.

75,000

175,000

1,100,000

11,000,000

16,000,000