There is a question that is asking me T=Temperature(Celsius) (t=minutes since poured in) where the function for finding temperature of coffee is $$ T=90(3^{-0.05t})$$ (initial coffee temp is 90) it wants me to find the amount of time (t) until the temperature has halved. How do i work backwards to get the time till initial temperature is half.
I just want to know any function or how to use this function to get the time until the initial temperature is halved already knowing the beginning value result of the function.
The equation you're solving is $$ 90\times 3^{-0.05t} = 45 $$ You can start by dividing both sides by $90$: $$ 3^{-0.05t} = \frac{1}{2} $$ Now take the natural logarithm on both sides: $$ \ln{(3^{-0.05t})} = -0.05t \ln{3} = \ln{(\frac{1}{2})} = -\ln{2} $$ Divide both sides by $-0.05 \ln{3}$ to get $$ t = \frac{\ln{2}}{0.05 \ln{3}} \approx 12.62 $$ Easy as can be!