Question

The sum of solutions of the equation log₂x.log₄x.log₆x = log₂x.log₄x + log₄x.log₆x + log₆x.log₂x is equal to​ a. 48 b. 49 c. 50 d. 47

Answer 1

Hi,

Answer A: 48

Explanation:

`log_2(x)*log_4(x)*log_6(x)=log_2(x)*log_4(x)+log_4(x)*log_6(x)+log_6(x)*log_2(x)\\\\\\(ln(x))/(ln(2)) *(ln(x))/(ln(4)) *(ln(x))/(ln(6)) =(ln(x))/(ln(2)) *(ln(x))/(ln(4)) +(ln(x))/(ln(4)) *(ln(x))/(ln(6)) +(ln(x))/(ln(6)) *(ln(x))/(ln(2))\\`

`\frac{{(ln(x))}^3}{ln(2)*ln(4)*ln(6)}=\frac{{(ln(x))}^2}{ln(2)*ln(4)}+\frac{{(ln(x))}^2}{ln(4)*ln(6)}+\frac{{(ln(x))}^2}{ln(2)*ln(6)}\\\\\\(ln(x))/(ln(2)*ln(4)*ln(6))=(1)/(ln(2)*ln(4))+(1)/(ln(4)*ln(6))+(1)/(ln(2)*ln(6))\\\\\\(ln(x))/(ln(2)*ln(4)*ln(6))=(ln(6)+ln(2)+ln(4))/(ln(2)*ln(4)*ln(6))\\\\\\ln(x)=ln(2*4*6)\\\\\\ln(x)=ln(48)\\\\\\\boxed{x=48}\\`

Answer A