Find the value of $x$ which satisfies ${log}_{x + 1} (x^{2} - 3 x + 1) = 1$

Open in App

Solution

Clearly $x + 1 > 0, x + 1 \neq 1$
${log}_{x + 1} (x^{2} - 3 x + 1) = 1$
$\Rightarrow x^{2} - 3 x + 1 = x + 1$
$\Rightarrow x = 4$
Ans: 4

Suggest Corrections

Similar questions

x, y, z

x + log (x + \sqrt{x^{2} + 1}) = y

y + log (y + \sqrt{y^{2} + 1}) = z

z + log (z + \sqrt{z^{2} + 1}) = x

{(1 - x)}^{2} + {(1 - y)}^{2} + {(1 - z)}^{2}

x

{log}_{x} (2 x^{2} - 3 x - 4) = 2

x^{2}

x

l o g (x + 1) = 2 l o g x

x

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

x

∣ ∣ ∣ \frac{x^{2}}{x - 1} ∣ ∣ ∣ \leq 1

Join BYJU'S Learning Program