How do you solve $\sqrt[5]{2 x + 1} + 5 = 9$ $root5(2x+1)+5=9$ ?

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Answer 1 · 2016-09-12T19:33:55

$x = 511.5$ $x = 511.5$

$\sqrt[5]{2 x + 1} + 5 = 9 \leftarrow$ $root5(2x+1)+5=9" "larr$ isolate the term with x

$\sqrt[5]{2 x + 1} = 9 - 5$ $root5(2x+1) = 9-5$

$\sqrt[5]{2 x + 1} = 4 \leftarrow$ $root5(2x+1) = 4 " "larr$ raise both sides to the power of 5

${(\sqrt[5]{2 x + 1})}^{5} = 4^{5} \leftarrow {(\sqrt[5]{x})}^{5} = x$ $(root5(2x+1))^5 = 4^5" " larr (root5 x)^5 = x$

$2 x + 1 = 1024$ $2x+1 = 1024$

$2 x = 1023$ $2x = 1023$

$x = 511.5$ $x = 511.5$

How do you solve 5√2x+1+5=9root5(2x+1)+5=9?