Algebra · real student question

Solve for x: 1/x³ − 1/(5 × 10⁻⁵)³ = 3.1100685 × 10¹¹.

Question

Solve for xx:

1x3−1(5×10−5)3=3.1100685×1011\frac{1}{x^3} - \frac{1}{\left(5\times 10^{-5}\right)^3} = 3.1100685\times 10^{11}

Step-by-step solution

  1. Evaluate the known constant first. Cube the base and the power of ten separately:

    (5×10−5)3=53×10−15=125×10−15=1.25×10−13\left(5\times 10^{-5}\right)^3 = 5^3 \times 10^{-15} = 125\times 10^{-15} = 1.25\times 10^{-13}

    The exponent is multiplied by 33, not added to 33 — 10−1510^{-15}, not 10−810^{-8}.

  2. Invert it. Reciprocating flips the mantissa and the sign of the exponent:

    11.25×10−13=0.8×1013=8×1012\frac{1}{1.25\times 10^{-13}} = 0.8 \times 10^{13} = 8\times 10^{12}

  3. Isolate 1/x³ by adding that constant to both sides.

    1x3=8×1012+3.1100685×1011\frac{1}{x^3} = 8\times 10^{12} + 3.1100685\times 10^{11}

    To add, put both terms on the same power of ten:

    =80×1011+3.1100685×1011=83.1100685×1011=8.31100685×1012= 80\times 10^{11} + 3.1100685\times 10^{11} = 83.1100685\times 10^{11} = 8.31100685\times 10^{12}

    Adding the mantissas without first matching exponents is the standard error here.

  4. Invert again to get x³.

    x3=18.31100685×1012=1.20322×10−13x^3 = \frac{1}{8.31100685\times 10^{12}} = 1.20322\times 10^{-13}

  5. Take the cube root. Split the mantissa from a power of ten whose exponent is a multiple of 33:

    x=120.322×10−153=120.3223×10−5=4.9368×10−5x = \sqrt[3]{120.322\times 10^{-15}} = \sqrt[3]{120.322}\times 10^{-5} = 4.9368\times 10^{-5}

    Rewriting 10−1310^{-13} as 120.322×10−15120.322\times10^{-15} is what makes the cube root of the power of ten exact.

  6. Check by substituting back. With x=4.9368×10−5x = 4.9368\times 10^{-5}:

    1x3=8.3110×1012,8.3110×1012−8×1012=3.110×1011 ✓\frac{1}{x^3} = 8.3110\times 10^{12}, \qquad 8.3110\times 10^{12} - 8\times 10^{12} = 3.110\times 10^{11} \ \checkmark

    Note how close xx is to 5×10−55\times 10^{-5}: the right-hand side is only about 3.9%3.9\% of 8×10128\times10^{12}, and because xx varies as the inverse cube root, that shifts xx by only about 1.3%1.3\%.

Answer

x≈4.9368×10−5x \approx 4.9368\times 10^{-5}

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