what is peak drug x level attained at steady state if 100mg is given intravenous injection every 6hrs. the patient ke=0.35/hr and Vd = 20L

The question is asking for the peak drug level at steady state. This can be calculated using the formula for the peak plasma concentration at steady state (Cpss max) for intermittent intravenous infusion, which is:

Cpss max = Dose / Volume of distribution * (1 - e^(-ketau)) / (1 - e^(-ketau*n))

where:

• Dose is the amount of drug given per infusion (100mg in this case)
• Volume of distribution (Vd) is the volume in which the drug is distributed (20L in this case)
• ke is the elimination rate constant (0.35/hr in this case)
• tau is the dosing interval (6 hours in this case)
• n is the number of doses given to reach steady state (this is typically assumed to be 4-5 half-lives, and the half-life (t1/2) can be calculated as 0.693/ke)

Let's calculate:

First, calculate the half-life: t1/2 = 0.693 / ke = 0.693 / 0.35 = 1.98 hours

Assume n = 5 half-lives to reach steady state, so n = 5 * 1.98 = 9.9 (round to 10 for simplicity)

Then, calculate Cpss max: Cpss max = 100mg / 20L * (1 - e^(-0.35/hr6hr)) / (1 - e^(-0.35/hr6hr*10))

This will give you the peak drug level at steady state.