calculus and exponential functions $0\leq t\leq3$, $t$ (measured in hours) $$G_1(t)=4e^{t/2}$$ Diabetic $$G_2(t)=8-4e^{-t/2}$$ Non diabetic- Relating to the question but the whole question isn't important, there are six parts and I cant seem to get two of them right. d) When does the diabetic patient have a glucose level of $10$? (G corresponds to glucose levels) My working out: $$G_1(t)=4e^{t/2}$$ Let $G_1=10$ $$10=4e^{t/2}$$ $$10/4=e^(t/2)$$ $$ln(10/4)=t/2$$ $$t= 2ln(10/4)$$ $$t \approx 1.8$$ (1 hour and 80 min so becomes two hours etc) however this isn't the answer :( it is actually 1 hour and 50 minutes! f) At which point in the experiment has the diabetic patient twice the glucose levels of the non diabetic patient? I know $$2G_1=G_2$$ $$2(4e^{t/2})=8-4e^{-t/2}$$ BUT i CANT SEEM TO FIND $t$ help please, thankyou in advance!

Part $1$ is good. See the comment of Alex Becker. $1.8$ hours is one hour and $0.8$ of another hour which consists of $60$ minutes. $0.8$ of $60$ is $0.8 \cdot 60 = 48$. Therefore $1.8$ hours is $1$ hour and $48$ minutes.

Part $2$

EDIT: You must set (!) $$ G_1 = 2G_2 $$ From $$ 4e^{\frac{t}2} = 2(8 - 4e^{\frac{-t}2}) $$ you get, by multiplying the equation by $e^{\frac{t}2}$, $$ 4e^t = 16 e^{\frac{t}2} - 8 $$ or $$ e^t - 4e^{\frac{t}2} + 2 = 0 $$ Now substitue $e^{\frac{t}2} = x$ and use $ (e^{\frac{t}2})^2 =e^{\frac{t}2} \cdot e^{\frac{t}2} = e^t$ to get $$ x^2 -4x + 2 =0 $$ Tell me if you can proceed from here. It looks like no real solution is there.

If we analyse a bit the function for diabetic patient we can see that it is exactly $4$ at $t=0$ and monotonically increases to $4e^{\frac32}=17.93$ at $t=3$.

On the other hand, the function for non-diabetic patient is exactly $4$ at $t=0$ and monotonically increases to $8-4e^{-\frac32}=7.11$ at $t=3$.