Mixture Problem 1. How much water should be added to 8 mL of 6% saline solution to reduce the concentration to 4%?

Amount of active ingredient is 0.04(8+x)= 0.32+0.4x

EQUATION:

active ingre. + active ingre. = active ingre

0.48 + 0 = 0.32 + 0.04x

0.16 = 0.04x

x=4ML (amount of water that must be added)

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

there is nothing below so i cannot

step-by-step explanation: