Question

The product of all positive real values of $x$ satisfying the equation $x^{\left(16\left(\log _5 x\right)^3-68 \log _5 x\right)}=5^{-16}$ is _____.

Answer 1

Taking log₅ on both sides
\((16(\log_5x)^3 -68(\log_5x))(\log_5x) = -16\)
Let ( log₅x)=t
16t⁴ - 68t² + 16 = 0
4t⁴ + 16t² - t²+4=0
(4t² - 1)(t²-4) = 0
\(t=±\frac{1}{2}\ or ±2\)
So log₅x = \(±\frac{1}{2}\ or ±2\)
\(⇒ x=5^{\frac{1}{2}} , 5^{-\frac{1}{2}} , 5^2, 5^{-2}\)
\(\therefore\) Product =  \((5)^{\frac{1}{2}-\frac{1}{2} +2-2}\)
\(= 5^0 = 1\)

So, the correct answer is 1.

Answer 2

Given equation $x^{\left(16\left(\log _5 x\right)^3-68 \log _5 x\right)}=5^{-16}$ Taking log₅ on both sides
\((16(\log_5x)^3 -68(\log_5x))(\log_5x) = -16\)
Let ( log₅x)=t
16t⁴ - 68t² + 16 = 0
4t⁴ + 16t² - t²+4=0//solving Quadratic Equation
(4t² - 1)(t²-4) = 0
\(t=±\frac{1}{2}\ or ±2\)
So log₅x = \(±\frac{1}{2}\ or ±2\)
\(⇒ x=5^{\frac{1}{2}} , 5^{-\frac{1}{2}} , 5^2, 5^{-2}\)
\(\rightarrow\)\(x=5^{\frac12-\frac12} or x=5^{2-2}\)
Therefore \(x=5^0 or x=5^0\)
\(x=1\) //Anything raised 0 is 1 

So, the correct answer is 1.