SOLUTION: The problem: Solve the equation. x^2(2x)-2^x=0 This is how I'm thinking it starts: x^2(2x)-2^x=0 2x^3-2^x=0 2(x^3-1^x)=0 x^3-1^x=0 3lnx=xln If this is right

Question 81739This question is from textbook College Algebra
: The problem: Solve the equation.
x^2(2x)-2^x=0
This is how I'm thinking it starts:
x^2(2x)-2^x=0
2x^3-2^x=0
2(x^3-1^x)=0
x^3-1^x=0
3lnx=xln
If this is right, I'm not sure where to take it at this point.
This question is from textbook College Algebra

Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
I like the way you started this one:
The problem: Solve the equation.
x^2(2x)-2^x=0
This is how I'm thinking it starts:
x^2(2x)-2^x=0
2x^3-2^x=0

However, the next step is not correct. You can't factor out the 2, and leave 1^x. If you did, it would be 2^(x-1), but you really don't want to do that. That gets really ugly. However, the whole problem is already ugly, since you have both an x as a base number and also as an exponent. Are you allowed to use a graphing calculator? If so, you might want to graph y=2x^3 -2^x, and determine where this graph crosses the x axis. The answers I got are x=1 and x=11.61301 approximately!! Maybe you should post this one again, and see if some of the other tutors know how to solve it algebraically.

R^2 at SCC

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