Answer
[tex]x_{1}=\frac{-3+33}{2}[/tex]
Explanation
Given
[tex]2\log_{\frac{1}{2}}(x+2)=\log_{\frac{1}{2}}(x+10)[/tex]
We can use the logarithmic properties to simplify our equation, where:
[tex]a\log_bx=\log_bx^a[/tex]
[tex]a^{\log_a(x)}=x[/tex]
Applying this rule to both sides of our problem given:
[tex]\frac{1}{2}^{\operatorname{\log}_{\frac{1}{2}}(x+2)2}=\frac{1}{2}^{\operatorname{\log}_{\frac{1}{2}}(x+10)}[/tex][tex](x+2)^2=(x+10)[/tex]
Solving the squared expression:
[tex]x^2+4x+4=x+10[/tex][tex]x^2+4x-x=10-4[/tex][tex]x^2+3x=6[/tex]
Setting the equation to 0:
[tex]x^2+3x-6=0[/tex]
Using a scientific calculator to get the result:
[tex]x_1=\frac{-3+\sqrt{33}}{2}[/tex][tex]x_1=\frac{-3-\sqrt{33}}{2}[/tex]
If we verify the solutions the correct one is the first.
