SOLUTION: can you please help me find the value of x? 2^5 = (1+0.23x)^3

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: can you please help me find the value of x? 2^5 = (1+0.23x)^3      Log On


   

Question 158861: can you please help me find the value of x?
2^5 = (1+0.23x)^3
Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!
2%5E5+=+%281%2B0.23x%29%5E3 -------------> basic eqn
32=%281%2B0.23x%29%5E3, raise the whole eqn by 1%2F3
32%5E%281%2F3%29=%281%2B0.23x%29%5E3%5E%281%2F3%29, cancel out the exponent on variable w/ "x"
3.1748=1%2B0.23x
3.1748-1=0.23x
2.1748=0.23x
cross%282.1748%299.45565%2Fcross%280.23%29=cross%280.23%29x%2Fcross%280.23%29
x=9.45565
check it out, go back basic eqn:
2%5E5=%281%2B0.23%289.45565%29%29%5E3
32=%281%2B2.1748%29%5E3
32=3.1748%5E3
32%3E=31.99999999
.32=32
Thank you,
Jojo