Question

Which equation is equivalent to 3logx+log2=log3x-log2

Answer 1

Answer:

x^2 = 3/4.

Step-by-step explanation:

3logx+log2=log3x-log2

Using the laws of logs:

logx^3 + log 2 = log 3x - log 2

log 2x^3 = log 3x/2

Removing the logs:

2x^3 = 3x/2

2x^2 = 3/2

x^2 =  3/4.

Answer 2

Step-by-step explanation:

$3\log x+\log2=\log3x-\log2\\\\\text{Domain:}\ x>0\\\\\text{Use}\\\\\log_ab^c=c\log_ab\\\\\log_ab+\log_ac=\log_a(bc)\\\\\log_ab-\log_ac=\log_a\left(\dfrac{b}{c}\right)\\===================\\\log x^3+\log2=\log\left(\dfrac{3x}{2}\right)\\\\\log(2x^3)=\log\left(\dfrac{3}{2}x\right)\iff2x^3=\dfrac{3}{2}x\qquad\text{subtract}\ \dfrac{3}{2}x\ \text{from both sides}\\\\2x^3-\dfrac{3}{2}x=0\qquad\text{multiply both sides by 2}\\\\4x^3-3x=0\qquad\text{use distributive property}\\\\x(4x^2-3)=0\iff x=0\ \vee\ 4x^2-3=0$

$x=0\notin\ \text{Domain}\\\\4x^2-3=0\qquad\text{add 3 to both sides}\\\\4x^2=3\qquad\text{divide both sides by 4}\\\\x^2=\dfrac{3}{4}\Rightarrow x=\pm\sqrt{\dfrac{3}{4}}\\\\x=\pm\dfrac{\sqrt3}{\sqrt4}\\\\x=-\dfrac{\sqrt3}{2}\notin \text{Domain}\\\\\boxed{x=\dfrac{\sqrt3}{2}}\in \text{Domain}$