lokiisnotdead

no problem! :)

hi otterstar!

so we know that \((-2)^4=16\) since -2*-2*-2*-2=16

since there are 4 roots, and we've already covered all the real ones, we have to look for complex numbers now.

we also know that \(i^4=1\), so that means that \((2i)^4=16\) since 2^4 already equals 16, and multiplying it by 1 will equal 16 too.

lastly, we know that \((-2i)^4=16\)  since we already know that -2^4 equals 16 (as we found earlier) and multiplying it by i^4=1 will equal 16 too. 

so the solutions are \(\boxed{2,-2,2i,-2i}\)

i hope this helped you! :)

oh wow! yay! :D

i think they are equal since if you simplify,

\(A=4^{5000}\cdot4^{5000}=2^{10000}\cdot2^{10000}=\boxed{2^{20000}}\)

\(B=2^{19999}+2^{19999}=2(2^{19999})=\boxed{2^{20000}}\)

I'm not sure if this is right, so please let me know if it is! :)

hi guest! 

i honestly don't know how to do this, but if you need quick answers, the answer is \(\boxed{x=\frac{17}{30}}\) (according to wolfram alpha)

ps how did you make the text colored? it looks really cool :)

there's 2 ways to solve this.

1. you can just experiment, and you can see that a=0, b=2, and c=7

2. we see that a+b+a+c=b+c

2a+b+c=b+c

2a=0

a=0

from there, you can find the other values :)

a+b+c=9

\(\sqrt[3]{x\sqrt{x}}=x^k\)

\(\sqrt[3]{x\cdot x^\frac{1}{2}}=x^k\)

\(\sqrt[3]{x^\frac{3}{2}}=x^k\)

\(x^{\frac{3}{2}\cdot\frac{1}{3}}=x^k\)

\(x^\frac{1}{2}=x^k\)

\(\boxed{k=\frac{1}{2}}\)

\(\)

\(\sqrt{7+\sqrt{4-\sqrt{3+x}}}=3\)

\(7+\sqrt{4-\sqrt{3+x}}=9\)     (squared both sides)

\(\sqrt{4-\sqrt{3+x}}=2\)             (subtracted 7 from both sides)

\(4-\sqrt{3+x}=4\)                 (square both sides again)

\(-\sqrt{3+x}=0\)                     (subtracted 4 from both sides)

\(\sqrt{3+x}=0\)                        (divided both sides by -1)

\(3+x=0\)                            (squared both sides again)

\(\boxed{x=-3}\)

\((x-11)(x+1)\)