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Solve the equation 5 x 1 - (2)/(x y) + _x y = -1 for x.

Mathematics

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Answer

x=1yorx=yx = \frac{1}{\sqrt{y}} \quad \text{or} \quad x = y

Step-by-step solution

5 steps

  1. Simplify the first term of the equation.

    We know that log⁡b1=0\log_b 1 = 0 for any valid base bb.
    Therefore, 5log⁡x1=5×0=05 \log_x 1 = 5 \times 0 = 0.

  2. Rewrite the equation with the simplified term.

    The equation becomes:

    0−2log⁡xy+log⁡xy=−10 - \frac{2}{\log_x y} + \log_x y = -1 −2log⁡xy+log⁡xy=−1-\frac{2}{\log_x y} + \log_x y = -1

  3. Introduce a substitution to simplify the equation.

    Let P=log⁡xyP = \log_x y.
    Substitute P into the equation:

    −2P+P=−1-\frac{2}{P} + P = -1

  4. Solve the equation for P.

    Multiply the entire equation by P (assuming P≠0P \neq 0):

    −2+P2=−P-2 + P^2 = -P

    Rearrange the terms to form a quadratic equation:

    P2+P−2=0P^2 + P - 2 = 0

    Factor the quadratic equation:

    (P+2)(P−1)=0(P+2)(P-1) = 0

    This gives two possible values for P:

    P=−2orP=1P = -2 \quad \text{or} \quad P = 1

  5. Substitute back P=log⁡xyP = \log_x y and solve for x in each case.

    Case 1: P=−2P = -2

    log⁡xy=−2\log_x y = -2

    By the definition of logarithms, x−2=yx^{-2} = y.

    1x2=y\frac{1}{x^2} = y x2=1yx^2 = \frac{1}{y}

    Taking the square root of both sides, and considering that the base of a logarithm must be positive (x>0x > 0):

    x=1yx = \sqrt{\frac{1}{y}} x=1yx = \frac{1}{\sqrt{y}}

    Case 2: P=1P = 1

    log⁡xy=1\log_x y = 1

    By the definition of logarithms, x1=yx^1 = y.

    x=yx = y

    The solutions for x are:

    x=1yorx=yx = \frac{1}{\sqrt{y}} \quad \text{or} \quad x = y
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