Answer:
x = 6
Step-by-step explanation:
Given : [tex]2\:log_2\:2\:+\:2\:log_26−\:log_2\:3x\:=\:3[/tex]
We have to solve the given expression [tex]2\:log_2\:2\:+\:2\:log_26−\:log_2\:3x\:=\:3[/tex]
Subtract [tex]2\log _2\left(2\right)+2\log _2\left(6\right)[/tex] both sides , we have,
[tex]2\:log_2\:2\:+\:2\:log_26-\:log_2\:3x-(2\log _2\left(2\right)+2\log _2\left(6\right)):=\:3-(2\log _2\left(2\right)+2\log _2\left(6\right))[/tex]
Simplify, we have,
[tex]\log _2\left(3x\right)=3-2\log _2\left(2\right)-2\log _2\left(6\right)[/tex]
Divide both side by -1, we have,
[tex]\frac{-\log _2\left(3x\right)}{-1}=\frac{3}{-1}-\frac{2\log _2\left(2\right)}{-1}-\frac{2\log _2\left(6\right)}{-1}[/tex]
Simplify, we have,
[tex]\log _2\left(3x\right)=-3+2\log _2\left(2\right)+2\log _2\left(6\right)[/tex]
Apply log rule, [tex]a=\log _b\left(b^a\right)[/tex]
[tex]2\log _2\left(6\right)-1=\log _2\left(2^{2\log _2\left(6\right)-1}\right)=\log _2\left(18\right)[/tex]
When log have same base,
[tex]\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\quad \Rightarrow \quad f\left(x\right)=g\left(x\right)[/tex]
[tex]\mathrm{For\:}\log _2\left(3x\right)=\log _2\left(18\right)\mathrm{,\:\quad solve\:}3x=18[/tex]
3x = 18
x = 6
