SOLUTION: I'm working with fraction exponents i.e. #roots. I know how to solve for the variable when it is an exponent, but this is a new lesson where I have to solve for the variable that i

Algebra ->  Sequences-and-series -> SOLUTION: I'm working with fraction exponents i.e. #roots. I know how to solve for the variable when it is an exponent, but this is a new lesson where I have to solve for the variable that i      Log On


   

Question 558170: I'm working with fraction exponents i.e. #roots. I know how to solve for the variable when it is an exponent, but this is a new lesson where I have to solve for the variable that is a base.
16d%5E%281%2F4%29=40
This what I've tried, but I've hit a dead end.
Problem: 16d%5E%281%2F4%29%7D=40
Step 1: %284d%5E2%29%5E%281%2F4%29=%284%5E3-24%29 or is it %284d%5E2%29%5E%281%2F4%29=%284%5E3%29-24
---I got same bases on both sides of the equation and had 4^(3-24) subsitute my value for 40, since (4^3)=64, but I would have to subtract 24 from either the exponent or from the whole side of the equation. Here's where I hit a wall and I don't know what to do after that since I tried to do both of my alternatives.
P.S. This is an excercise to get our feet wet for tomorrow's lesson.
Found 3 solutions by KMST, edjones, mananth:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You don't want addition and/or subtraction. Stick to multiplication, division and powers.
First, division:
16d%5E%281%2F4%29=40 --> d%5E%281%2F4%29=40%2F16 --> d%5E%281%2F4%29=5%2F2
Now, powers:
d%5E%281%2F4%29=5%2F2 --> %28d%5E%281%2F4%29%29%5E4=%285%2F2%29%5E4 --> d=625%2F16

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
16d^(1/4)=40
d^(1/4)=5/2
d=(5/2)^4=625/16 Take each side to the 4th power.
.
Ed

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
16d%5E%281%2F4%29=40


raise to power 4 on both sides


%2816%5E4%2Ad%5E%281%2F4%29%29%5E4+=+40+%5E4
16^4*d= 40^4
d= 40^4/16^4
d=625/16