Questions on Algebra: Negative and Fractional exponents answered by real tutors!

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Question 158861: can you please help me find the value of x?
2^5 = (1+0.23x)^3: can you please help me find the value of x?
2^5 = (1+0.23x)^3
Answer by jojo14344(1022) About Me  (Show Source):
You can put this solution on YOUR website!
2^5 = (1+0.23x)^3 -------------> basic eqn
32=(1+0.23x)^3, raise the whole eqn by 1/3
32^(1/3)=(1+0.23x)^3^(1/3), cancel out the exponent on variable w/ "x"
3.1748=1+0.23x
3.1748-1=0.23x
2.1748=0.23x
cross(2.1748)9.45565/cross(0.23)=cross(0.23)x/cross(0.23)
x=9.45565
check it out, go back basic eqn:
2^5=(1+0.23(9.45565))^3
32=(1+2.1748)^3
32=3.1748^3
32>=31.99999999
.32=32
Thank you,
Jojo