Question

The solution set of 2x^3/4+1=17 is?
Fractions like that confuse me ._.

Answer 1

Is it \[\LARGE 2x^{\frac{3}{4}}+1=17\] or is it \[\LARGE \frac{2x^{3}}{4}+1=17\] ???

Answer 2

The first one jim

Answer 3

\[2x^{\frac{3}{4}}+1=17\]
\[2x^{\frac{3}{4}}=16\]
\[x^{\frac{3}{4}}=8\] is a start

Answer 4

then
\[x=8^{\frac{4}{3}}=\sqrt[3]{8^4}=2^4=16\]

Answer 5

\[\LARGE 2x^{\frac{3}{4}}+1=17\]


\[\LARGE 2\sqrt[4]{x^3}+1=17\]


\[\LARGE 2\sqrt[4]{x^3}=16\]


\[\LARGE \sqrt[4]{x^3}=8\]


\[\LARGE x^3=8^4\]


\[\LARGE x^3=(2^3)^4\]


\[\LARGE x^3=2^{3*4}\]


\[\LARGE x^3=2^{4*3}\]


\[\LARGE x^3=(2^4)^3\]


\[\LARGE x=\sqrt[3]{(2^4)^3}\]


\[\LARGE x=2^4\]


\[\LARGE x=16\]

So the answer is \[\LARGE x=16\]

Answer 6

Ahhhh this makes so much sense now. Thank you!