Question

Please show me how to solve this and check for extraneous solutions. 3x^4/3-7=41

Answer 1

Do you mean\[\frac{ 3x^4 }{ 3 } - 7 = 41\]Or is (3-7) in the denominator?

Answer 2

No, 3 is the denominator for the 4 only. Its part of that exponent

Answer 3

Okay. So can you start off by isolating the 3x^4? Can you do that by yourself, or do you need help with that?

Answer 4

I need help because I don't know how to do 3x^4/3

Answer 5

Wait...is it \[3x^{4/3}\]or what I wrote up there?

Answer 6

Yes. Its the new one u just wrote

Answer 7

Okay. Now I understand why you're having problems.

What you need to first do is isolate the x, so we have\[x^{4/3} = 16\]Now we need to figure out what to do with the exponent. What you need to do is raise both sides to the 3/4 power. This will give us just an x on the left side.\[(x^{4/3})^{3/4} = x = 16^{3/4} = \sqrt[4]{16^3} = 2^3 = 8\]So x=8. Do you need me to go over the last part?

Answer 8

@Dee1516?

Answer 9

And thanks for the medal, @DebbieG. I appreciate it.

Answer 10

No problem :)

Answer 11

Thank you so much! @oaktree

Answer 12

No problem. I'm glad I could help. :)

Answer 13

Can you help me with this one as well 4(x-3)^3/2+12=120

Answer 14

@oaktree

Answer 15

Already here!

Same idea. Subtract 12 and divide by 4 to get\[(x-3)^{3/2} = 27\]Then raise both sides to the 2/3 power and you get\[x-3 = 27^{2/3} = ((27)^{1/3})^2 = 3^2 = 9\]But don't forget we still need to isolate the x! So we add 3 and get x=12.

Answer 16

Good?

Answer 17

Yes! Thank you so much!!!!

Answer 18

Great. Just fan me or write a testimonial or something and that way you can send me a call next time you have a problem. Glad to help!

Answer 19

Okay!