Question

(1/3)x^(-2/3) - (1/5)x^(-4/5) = 0

Answer 1

might try to factor out x^(-4/5) to begin?

Answer 2

\(\large \bf \cfrac{1}{3}x^{-\frac{2}{3}}-\cfrac{1}{5}x^{-\frac{4}{5}}=0\quad ?\)

Answer 3

yes

Answer 4

came from taking the derivative of a function

Answer 5

k

Answer 6

got to a point where the equation is x= (125/27) * x^(15/4) but may have gone wrong

Answer 7

\(\Large {
\cfrac{1}{3}x^{-\frac{2}{3}}-\cfrac{1}{5}x^{-\frac{4}{5}}=0\implies \cfrac{x^{-\frac{2}{3}}}{3}-\cfrac{x^{-\frac{4}{5}}}{5}=0
\\ \quad \\
\cfrac{x^{-\frac{2}{3}}}{3}=\cfrac{x^{-\frac{4}{5}}}{5}\implies \cfrac{x^{-\frac{2}{3}}}{x^{-\frac{4}{5}}}=\cfrac{3}{5}\implies x^{-\frac{2}{3}+\frac{4}{5}}=\cfrac{3}{5}
\\ \quad \\
x^{\frac{2}{15}}=\cfrac{3}{5}\implies \left(x^{\cancel{ \frac{2}{15}}}\right)^{\cancel{ \frac{15}{2}}}=\left(\cfrac{3}{5}\right)^{\frac{15}{2}}
}\)

thus far it looks like

Answer 8

k

Answer 9

would that be plus/minus, since we took the square root of a square at some point?

Answer 10

hmm
well... can be either I gather... we started off with a rational exponent though
but since is square root... could be either, depending on context