Question

How do you algebraically solve the equation (2^-x)-6=0?

Answer 1

Add 6 to both sides.
Take the natural log of both sides:\[\ln(2^{-x} ) = \ln(6)\]
You can bring the -x out front:\[-xln2 = \ln6\]
Then divide both sides by -ln2:\[x = \frac{-\ln6}{\ln2}\]

Answer 2

Thanks. These types of equations always trip me up..