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Quadratic Equation and Inequalities question

2022 · Shift 2 · Q22
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  5. /2022 · Shift 2 · Q22

Quadratic Equation and Inequalities question

2022 · Shift 2 · Q22

JEE AdvancedMathematicsQuadratic Equation and InequalitiesNumerical+3 / −1
The product of all positive real values of xxx satisfying the equation x(16(log⁡5x)3−68log⁡5x)=5−16x^{\left(16\left(\log _{5} x\right)^{3}-68 \log _{5} x\right)}=5^{-16}x(16(log5​x)3−68log5​x)=5−16 is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 1

Let t=log⁡5x.t=\log_5 x.t=log5​x. Then x=5t,x(16(log⁡5x)3−68log⁡5x)=(5t)(16t3−68t)=5t(16t3−68t)=516t4−68t2.x=5^t,\qquad x^{\left(16(\log_5 x)^3-68\log_5 x\right)}=(5^t)^{\left(16t^3-68t\right)}=5^{t(16t^3-68t)}=5^{16t^4-68t^2}.x=5t,x(16(log5​x)3−68log5​x)=(5t)(16t3−68t)=5t(16t3−68t)=516t4−68t2.

So the given equation becomes 516t4−68t2=5−16.5^{16t^4-68t^2}=5^{-16}.516t4−68t2=5−16. Since the bases are equal and positive, 16t4−68t2=−16.16t^4-68t^2=-16.16t4−68t2=−16.

1. Form a quadratic in t2t^2t2

Bring all terms to one side: 16t4−68t2+16=0.16t^4-68t^2+16=0.16t4−68t2+16=0. Divide by 444: 4t4−17t2+4=0.4t^4-17t^2+4=0.4t4−17t2+4=0. Let y=t2.y=t^2.y=t2. Then 4y2−17y+4=0.4y^2-17y+4=0.4y2−17y+4=0.

Solve this quadratic: 4y2−17y+4=(4y−1)(y−4)=0.4y^2-17y+4=(4y-1)(y-4)=0.4y2−17y+4=(4y−1)(y−4)=0. Hence, y=14ory=4.y=\frac14 \quad \text{or} \quad y=4.y=41​ory=4. So, t2=14ort2=4.t^2=\frac14 \quad \text{or} \quad t^2=4.t2=41​ort2=4. Therefore, t=±12, ±2.t=\pm\frac12,\ \pm 2.t=±21​, ±2.

2. Convert back to xxx

Since x=5tx=5^tx=5t, the positive real solutions are x=51/2, 5−1/2, 52, 5−2.x=5^{1/2},\ 5^{-1/2},\ 5^2,\ 5^{-2}.x=51/2, 5−1/2, 52, 5−2. That is, x=5, 15, 25, 125.x=\sqrt5,\ \frac1{\sqrt5},\ 25,\ \frac1{25}.x=5​, 5​1​, 25, 251​.

3. Find their product

The product is 5⋅15⋅25⋅125=1.\sqrt5\cdot \frac1{\sqrt5}\cdot 25\cdot \frac1{25}=1.5​⋅5​1​⋅25⋅251​=1.

Therefore, the required product is 1.\boxed{1}.1​.

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