The sum of all the values of $x$ satisfying the equation $({log}_{2} x) ({log}_{3} x) = {log}_{3} (x^{3}) + {log}_{2} (x^{2}) - 6$ is

Open in App

Solution

Clearly $x > 0$
Put ${log}_{2} x = y, {log}_{3} x = z$ to obtain
$y z = 3 z + 2 y - 6 [∵ log a^{m} = m log a]$
$\Rightarrow (y - 3) (z - 2) = 0$
$\Rightarrow x = 8, 9$
Therefore, sum of roots $= 17$
Ans: 17

Suggest Corrections

Similar questions

x^{{log}_{3} x^{2} + {({log}_{3} x)}^{2} - 10} = \frac{1}{x^{2}}

2 {log}_{2} {log}_{2} x + {log}_{1 / 2} {log}_{2} (2 \sqrt{2} x) = 1

x

{log}_{2} (x^{2} - 3) - {log}_{2} (6 x - 10) + 1 = 0

x

{log}_{2} (9 - 2^{x}) = 3 - x

x^{(\frac{3}{4} {({log}_{2} x)}^{2} + {log}_{2} x - \frac{5}{4})} = \sqrt{2}

Join BYJU'S Learning Program