Exponents

x^1/4 +2=4

Please Help[/img]

\(\displaystyle \L
\begin{array}{l}
\sqrt[4]{x} + 2 = 4 \\
\sqrt[4]{x} = 2 \\
x = ? \\
\end{array}\)

Could you tell me if I came to the right conclusion? My answer is
x=2^4
x=16

To check any "solving" exercise, plug your solution back into the original equation:


. . . . .\(\displaystyle x^{\frac{1}{4}}\,+\,2\,=\,4\)


. . . . .\(\displaystyle (16)^{\frac{1}{4}}\,+\,2\,\begin{array}{c}?\\=\end{array}\,4\)


. . . . .\(\displaystyle 2\,+\,2\,\begin{array}{c}?\\=\end{array}\,4\)


. . . . .\(\displaystyle 4\,=\,4\)


Your solution "checks". Good work!

Eliz.

Thanks!