The number of the integer roots of the equation ${log}_{(x^{2} - 1)} (x^{3} + 6) = {log}_{(x^{2} - 1)} ({2 x}^{2} + 5 x)$ , is:

Open in App

Solution

This equation is equivalent to the system
$⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {2 x}^{2} + 5 x > 0 x^{2} - 1 > 0 x^{2} - 1 \neq 1 x^{3} + 6 = {2 x}^{2} + 5 x \end{matrix}$ $\Rightarrow ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x \in (- \infty, - \frac{5}{2}) \cup (0, \infty) | x | > 1 x \neq \pm \sqrt{2} x = {- 2, 1, 3} \end{matrix}$
Hence, on intersection of these values of $x$ , we get $x = 3$ is the only root of the original equation.

Suggest Corrections

Similar questions

x

{log}_{x^{2} - 1} (x^{3} + 6) = {log}_{x^{2} - 1} (4 x^{2} - 4)

y = tan [log (x^{2})]

\frac{d y}{d x}

{log}_{x^{2} - 6 x + 8} ({log}_{{2 x}^{2} - 2 x - 8} (x^{2} + 5 x)) = 0

x

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

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

Join BYJU'S Learning Program