To calculate the value of the half life we proceed as follows; a/2=ae^(-2.360t) the a(s) will cancel and we shall remain with: 1/2=e^-2.360t introducing the natural logs we get: ln (1/2)=ln e^(-2.360t) ln(1/2)=-2.360t hence; t=(ln 0.5)/(-2.360) t=0.294 the answer is t=0.294
To solve using completing square method we proceed as follows: x^2-10x+8=0 x^2-10x=-8 but c=(b/2)^2 c=(10/2)^2=25 thus we can add this in our expression to get x^2-10x+25=8+25 factorizing the LHS we get: (x-5)(x-5)=33 (x-5)^2=33 getting the square roots of both sides we have: x-5=+/-√33 x=5+/-√33
