Question

please solve this JEE advance level problem?​

Answer 1

Answer:

x=30

Step-by-step explanation:

Given,

$log_{2}(x) . log_{3}(x) . log_{5}(x) = log_{2}(x). log_{3}(x) + log_{3}(x). log_{5}(x) + log_{5}(x). log_{2}(x)$

$\frac{ ln(x) }{ ln(2) } . \frac{ ln(x) }{ ln(3) }. \frac{ ln(x) }{ ln(5) } = \frac{ ln(x) }{ ln(2) } . \frac{ ln(x) }{ ln(3) } + \frac{ ln(x) }{ ln(3) } . \frac{ ln(x) }{ ln(5) } + \frac{ ln(x) }{ ln(5) }. \frac{ ln(x) }{ ln(2) }$

as we know that,

$log_{a}(b) = \frac{ log(a) }{ log(b) }$

$\frac{ { (ln(x) )}^{3} }{ ln(2). ln(3). ln(5) } = ( ln(x))^{2} ( \frac{ ln(2) + ln(3) + ln(5) }{ ln(2). ln(3). ln(5) } )$

on simplyfing,

$ln(x) = ln(2) + ln(3) + ln(5)$

As we know,

$log(a.b.c) = log(a) + log(b) + log(c)$

so, we have,

$ln(x) = ln(2.3.5)$

$ln(x) = ln(30)$

$x = 30$

Hope this will help you..!