solve for the value of x in the equation x= 2^{x+6}

solve for the value of x in the equation x= 2^{x+6}

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If there was a solution, it would have to be positive, since RHS is always positive.

The average slope of RHS from x = -1 to 0 is 32 and the slope strictly increases as x increases. Thus the slope of RHS is greater than 32 at all x > 0 so the RHS increases faster than x and RHS > x at x=0. Thus there can't be any intersection.

If you are taking calculus, then there is another way:

Let y = 2^{x+6} - x. Then dy/dx = 2^{x+6}*ln 2 - 1. Setting dy/dx = 0 gives.

2^{x+6}*ln 2 = 1

2^{x+6} = 1/ln 2

x+6 = -log_{2} ln 2

x = -6 - log_{2} ln 2

At this point, y = (1/ln 2) + 6 + log_{2} ln 2 > 0.

d^{2}y/dx^{2} = 2^{x+6} (ln 2)^{2} > 0 for all x so y = (1/ln 2) + 6 + log_{2} ln 2 is a global minimum and so 2^{x+6} > x for all real x. Thus there is no solution.

^{6}, which is clearly not true.

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