Been too many years since algebra...

[pgc01]

Hi Greg

I don't know what's the equation for, so maybe it's not relevant but this equation may have more than 1 solution.

If it's not relevent please discard the rest.

A simple way to determine the number of solutions is to solve the equation grafically.

• k = x · (ba<SUP>xc</SUP> + d)

• k/x - d = ba<SUP>xc</SUP>

The right side curve is an exponential. The left side is a hyperbola (k/x) symetrical to the origin to which is applied a translation of value -d.

This way it's easy to know what to look for.

Ex.: a=1.5, b=1.2, c=2, k=1

• 1/x - d = 1.2 × 1.5<SUP>2x</SUP>

So it depends on d. If d is positive the hyperbola is pushed down and we have 1 solution. Ex.: d=1

• 1/x - 1 = 1.2 × 1.5<SUP>2x</SUP>

x = 0.3798

BUT, if If d is negative the hyperbola is pushed up and we have the exponential is quadrants 1 and 2 and the hyperbola in quadrants 1, 2 and 3. This gives 2 solutions. Ex.: d=-1

• 1/x + 1 = 1.2 × 1.5<SUP>2x</SUP>

Starting with the x cell empty Goal seek gives the answer

x = 0.7866

but now I know that there's another solution, and I get

x = -1.5309

Pedro