JEE Advance - Mathematics (2022 - Paper 2 Online - No. 4)
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 __________.
$$ x^{\left(16\left(\log _{5} x\right)^{3}-68 \log _{5} x\right)}=5^{-16} $$
is __________.
Answer
1
Explanation
Taking log to the base 5 on both sides
$\left(16\left(\log _{5} x\right)^{3}-68\left(\log _{5} x\right)\right)\left(\log _{5} x\right)=-16$
Let $\left(\log _{5} x\right)=t$
$16 t^{4}-68 t^{2}+16=0$
$$ \Rightarrow $$ $4 t^{4}-16 t^{2}-t^{2}+4=0$
$$ \Rightarrow $$ $\left(4 t^{2}-1\right)\left(t^{2}-4\right)=0$
$$ \Rightarrow $$ $t=\pm \frac{1}{2}, \pm 2$
So $\log _{5} x=\pm \frac{1}{2}$ or $\pm 2$
$\Rightarrow x=5^{\frac{1}{2}}, 5^{\frac{-1}{2}}, 5^{2}, 5^{-2}$
$\left(16\left(\log _{5} x\right)^{3}-68\left(\log _{5} x\right)\right)\left(\log _{5} x\right)=-16$
Let $\left(\log _{5} x\right)=t$
$16 t^{4}-68 t^{2}+16=0$
$$ \Rightarrow $$ $4 t^{4}-16 t^{2}-t^{2}+4=0$
$$ \Rightarrow $$ $\left(4 t^{2}-1\right)\left(t^{2}-4\right)=0$
$$ \Rightarrow $$ $t=\pm \frac{1}{2}, \pm 2$
So $\log _{5} x=\pm \frac{1}{2}$ or $\pm 2$
$\Rightarrow x=5^{\frac{1}{2}}, 5^{\frac{-1}{2}}, 5^{2}, 5^{-2}$
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