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Difficulty: Very hardExponents, Roots, and Scientific Notation

In astrophysics, the energy intensities of two different cosmic signals are modeled. The primary signal has an intensity of I1=x⋅y23I_1 = \sqrt[3]{x \cdot y^2} watts per square meter, where x=4.0×104x = 4.0 \times 10^4 and y=5.0×103y = 5.0 \times 10^3. The secondary signal has an intensity of I2=z3⋅w−12.0×10−2I_2 = \frac{\sqrt{z^3 \cdot w^{-1}}}{2.0 \times 10^{-2}} watts per square meter, where z=1.6×10−3z = 1.6 \times 10^{-3} and w=2.5×10−7w = 2.5 \times 10^{-7}. What is the ratio of the primary signal intensity to the secondary signal intensity, expressed in scientific notation?

  1. A
    6.25×1026.25 \times 10^2
  2. B
    6.4×10−46.4 \times 10^{-4}
  3. C
    1.5625×1021.5625 \times 10^2
  4. 1.5625×1031.5625 \times 10^3Answer
  5. E
    1.5625×1071.5625 \times 10^7

Answer

The correct ratio of the primary signal intensity to the secondary signal intensity is 1.5625×1031.5625 \times 10^3.
To find the correct ratio of the primary signal intensity to the secondary signal intensity, first compute the value of both intensities. The primary signal intensity is I1=x⋅y23=4.0×104⋅(5.0×103)23=1.0×10123=1.0×104=10,000I_1 = \sqrt[3]{x \cdot y^2} = \sqrt[3]{4.0 \times 10^4 \cdot (5.0 \times 10^3)^2} = \sqrt[3]{1.0 \times 10^{12}} = 1.0 \times 10^4 = 10,000. The secondary signal intensity is evaluated by simplifying the numerator term: z3⋅w−1=(1.6×10−3)3⋅(2.5×10−7)−1=4.096×10−9⋅4.0×106=1.6384×10−2z^3 \cdot w^{-1} = (1.6 \times 10^{-3})^3 \cdot (2.5 \times 10^{-7})^{-1} = 4.096 \times 10^{-9} \cdot 4.0 \times 10^6 = 1.6384 \times 10^{-2}. Taking the square root gives 1.28×10−11.28 \times 10^{-1}. Dividing this by the denominator yields I2=1.28×10−12.0×10−2=6.4I_2 = \frac{1.28 \times 10^{-1}}{2.0 \times 10^{-2}} = 6.4. Finally, the ratio of the primary intensity to the secondary intensity is 100006.4=1562.5\frac{10000}{6.4} = 1562.5, which is expressed in standard scientific notation as 1.5625×1031.5625 \times 10^3.

Step-by-Step Solution

1
Calculate the squared value of the variable y to evaluate the primary signal intensity.
y2=(5.0×103)2=5.02×(103)2=25×106=2.5×107y^2 = (5.0 \times 10^3)^2 = 5.0^2 \times (10^3)^2 = 25 \times 10^6 = 2.5 \times 10^7
Squaring a term in scientific notation requires squaring the coefficient and multiplying the exponent of 10 by 2.
2
Multiply the variable x by the calculated value of y^2 and take the cube root to find the primary intensity.
x⋅y2=(4.0×104)×(2.5×107)=10×1011=1.0×1012x \cdot y^2 = (4.0 \times 10^4) \times (2.5 \times 10^7) = 10 \times 10^{11} = 1.0 \times 10^{12}; therefore, I1=1.0×10123=1.0×104=10,000I_1 = \sqrt[3]{1.0 \times 10^{12}} = 1.0 \times 10^4 = 10,000
To multiply numbers in scientific notation, multiply their coefficients and add their exponents. To find the cube root, raise both parts to the 1/3 power.
3
Evaluate the expressions for z^3 and w^{-1} in the numerator of the secondary signal intensity.
z3=(1.6×10−3)3=4.096×10−9z^3 = (1.6 \times 10^{-3})^3 = 4.096 \times 10^{-9} and w−1=(2.5×10−7)−1=0.4×107=4.0×106w^{-1} = (2.5 \times 10^{-7})^{-1} = 0.4 \times 10^7 = 4.0 \times 10^6
To cube z, cube the coefficient and multiply the exponent by 3. To find the reciprocal of w, invert the coefficient and negate the exponent of 10, then rewrite in standard form.
4
Multiply z^3 by w^{-1} and take the square root to find the numerator of the secondary intensity.
z3⋅w−1=16.384×10−3=1.6384×10−2z^3 \cdot w^{-1} = 16.384 \times 10^{-3} = 1.6384 \times 10^{-2}; therefore, z3⋅w−1=1.28×10−1=0.128\sqrt{z^3 \cdot w^{-1}} = 1.28 \times 10^{-1} = 0.128
Multiply the coefficients and add the exponents. Take the square root of the resulting coefficient and divide the exponent of 10 by 2.
5
Divide the simplified numerator by the denominator of the secondary signal intensity.
I2=1.28×10−12.0×10−2=0.64×101=6.4I_2 = \frac{1.28 \times 10^{-1}}{2.0 \times 10^{-2}} = 0.64 \times 10^1 = 6.4
Divide the coefficients and subtract the exponent in the denominator from the exponent in the numerator.
6
Determine the ratio of the primary intensity to the secondary intensity.
I1I2=100006.4=1562.5=1.5625×103\frac{I_1}{I_2} = \frac{10000}{6.4} = 1562.5 = 1.5625 \times 10^3
Divide the primary signal intensity value by the secondary signal intensity value, then convert the result to standard scientific notation.

Key Concept

Simplifying algebraic operations on terms involving powers, roots, and scientific notation.
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Topics

Exponents, Roots, and Scientific Notation

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