To find the domain of a function, look for the restrictions over the functions involved in it.
[tex]\text{Let }f(x)=\log _3(\sqrt[]{x}+1)[/tex]The log functions are defined whenever their argument is greater than 0. This means:
[tex]\sqrt[]{x}+1>0[/tex]Since 1>0 and the square root of x is always greater than or equal to 0, then this does not restrict the domain of the function.
Nevertheless, the square root of x requires that:
[tex]x\ge0[/tex]This is the only restriction over the variable x.
Therefore, the domain of the function, is:
[tex]0\leq x[/tex]Next, solve the equation:
[tex]\log _3(\sqrt[]{x}+1)=1[/tex]Notice that given this equation, then 3 to the power of each side of the equation should be equal:
[tex]3^{\log _3(\sqrt[]{x}+1)}=3^1[/tex]Use the fact that:
[tex]3^{\log _3(a)}=a[/tex]to simplify the equation:
[tex]\begin{gathered} 3^{\log _3(\sqrt[]{x}+1)}=3^1 \\ \Rightarrow \\ \sqrt[]{x}+1=3^1 \end{gathered}[/tex]Simplify the power 3^1:
[tex]\sqrt[]{x}+1=3[/tex]Substract 1 from both sides of the equation:
[tex]\sqrt[]{x}=2[/tex]Square both sides of the equation:
[tex](\sqrt[]{x})^2=2^2[/tex]Simplify the powers:
[tex]x=4[/tex]Check the answer by plugging in x=4 into the equation:
[tex]\begin{gathered} \log _3(\sqrt[]{x}+1)=1 \\ \Rightarrow \\ \log _3(\sqrt[]{4}+1)=1 \\ \Rightarrow \\ \log _3(2+1)=1 \\ \Rightarrow \\ \log _3(3)=1 \\ \Rightarrow \\ 1=1 \end{gathered}[/tex]Since we got an identity, the answer x=4 is correct.
