SOLUTION: can you please help me find x? 2X10^-5 = (1+0.23x)^-3 Thank you in advance.

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: can you please help me find x? 2X10^-5 = (1+0.23x)^-3 Thank you in advance.      Log On


   

Question 159050: can you please help me find x?
2X10^-5 = (1+0.23x)^-3
Thank you in advance.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
2%2A10%5E%28-5%29+=+%281+%2B+.23x%29%5E%28-3%29
multiply both sides by this expression:
%2810%5E5%29%2A%281+%2B+.23x%29%5E3

This reduces to
2%2A%281+%2B+.23x%29%5E3+=+10%5E5
%281+%2B+.23x%29%5E3+=+.5%2A10%5E5
Take the cube root of both sides
1+%2B+.23x+=+%2850000%29%5E%281%2F3%29
1+%2B+.23x+=+36.84
.23x+=+36.84+-+1
.23x+=+35.84
x+=+155.83 answer
Hope I got it right