Show that x = 2 is the only root of the equation
$9^{l o g_{3} [l o g_{2} x]} = l o g_{2} x - (l o g_{2} x)^{2} + 1$

Open in App

Solution

$l o g_{3} (l o g_{2} x)$ is defined only when $l o g_{2} x = t$ is +ive, i,e., $l o g_{2} x > 0 = l o g_{2} 1$
$∴$ x > 1
Also $a^{l o g_{a}^{n}} \Rightarrow 3^{2 l o g_{3} (t)} = 3^{l o g_{3} (t^{2})} = t^{2}$
$∴ t^{2} = t - t^{2} + 1$
or $2 t^{2} - t - 1 = 0$ or (2t + 1)(t - 1) = 0
$∴$ t = 1 only $(- \frac{1}{2} r e j e c t e d a s t i s + i v e)$
$∴ l o g_{2} x = 1 ∴ x = 2$
Thus there is only one root 2.

Suggest Corrections

Similar questions

x

g^{l o g_{3} (l o g_{2} x)} = l o g_{2} x - (l o g_{2} x)^{2} + 1

x^{\frac{2}{3} [(l o g_{2} x)^{2} + l o g_{2} x - 5 / 4]} = \sqrt{2}

3 x^{2} - 2 x^{3} = {log}_{2} (x^{2} + 1) - {log}_{2} x

| {log}_{2} x + 1 | + ∣ ∣ 1 - ({log}_{2} x)^{2} ∣ ∣ = ∣ ∣ {log}_{2} x + ({log}_{2} x)^{2} ∣ ∣,

x

{λ} \cup [μ, \infty) .

3^{{log}_{3} ({log}_{2} x)^{2}} = {log}_{2} x - ({log}_{2} x)^{2} + 1

x

Join BYJU'S Learning Program