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Algebra · real student question

Solve the inequality (2/5)^x < sqrt((4/25)^((3x + 2)/x)).

Question

Solve the inequality

(25)x<(425)3x+2x\left(\frac{2}{5}\right)^{x}<\sqrt{\left(\frac{4}{25}\right)^{\frac{3x+2}{x}}}

Step-by-step solution

  1. State the domain first. The exponent 3x+2x\frac{3x+2}{x} is undefined at x=0x=0, so x≠0x\ne 0 is a standing restriction. Everything else is defined for all real xx, since a positive base raised to any real power is positive.

  2. Write both sides with the same base. Notice 425=(25)2\frac{4}{25}=\left(\frac{2}{5}\right)^2, so the square root cancels the square:

    (425)3x+2x=[(25)2⋅3x+2x]1/2=(25)3x+2x\sqrt{\left(\frac{4}{25}\right)^{\frac{3x+2}{x}}}=\left[\left(\tfrac{2}{5}\right)^{2\cdot\frac{3x+2}{x}}\right]^{1/2}=\left(\frac{2}{5}\right)^{\frac{3x+2}{x}}

    The inequality becomes a pure comparison of exponents on the same base:

    (25)x<(25)3x+2x\left(\frac{2}{5}\right)^{x}<\left(\frac{2}{5}\right)^{\frac{3x+2}{x}}

  3. Flip the inequality because the base is less than 1. The function t↦(25)tt\mapsto\left(\frac25\right)^t is decreasing, so a smaller value corresponds to a larger exponent:

    x>3x+2xx>\frac{3x+2}{x}

    Forgetting this reversal is the single most common error in exponential inequalities with a fractional base.

  4. Move everything to one side — never multiply by xx. Multiplying across by xx would require splitting into cases for the sign of xx; subtracting is safer:

    x−3x+2x>0  ⟺  x2−3x−2x>0x-\frac{3x+2}{x}>0\;\Longleftrightarrow\;\frac{x^2-3x-2}{x}>0

  5. Find the critical points and build a sign chart. The numerator vanishes when x2−3x−2=0x^2-3x-2=0, i.e.

    x=3±9+82=3±172≈−0.5616 and 3.5616x=\frac{3\pm\sqrt{9+8}}{2}=\frac{3\pm\sqrt{17}}{2}\approx-0.5616\ \text{and}\ 3.5616

    Together with the zero of the denominator, the critical points in order are 3−172\frac{3-\sqrt{17}}{2}, 00, 3+172\frac{3+\sqrt{17}}{2}. Testing one point in each of the four intervals gives the signs −,  +,  −,  +-,\;+,\;-,\;+.

  6. Write the solution set and spot-check it. The quotient is positive on

    x∈(3−172, 0)∪(3+172, +∞)x\in\left(\frac{3-\sqrt{17}}{2},\,0\right)\cup\left(\frac{3+\sqrt{17}}{2},\,+\infty\right)

    Check x=−0.4x=-0.4: left side 0.4−0.4≈1.440.4^{-0.4}\approx1.44, right side 0.4(3(−0.4)+2)/(−0.4)=0.4−2=6.250.4^{(3(-0.4)+2)/(-0.4)}=0.4^{-2}=6.25 — the inequality holds. Check x=1x=1: 0.4<0.45=0.010.4<0.4^{5}=0.01 is false, as the chart predicts.

Answer

x∈(3−172, 0)∪(3+172, +∞)x\in\left(\frac{3-\sqrt{17}}{2},\,0\right)\cup\left(\frac{3+\sqrt{17}}{2},\,+\infty\right)

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