How do you isolate "x" when it is in an equation like this: 5.21 = 4.74 + [log (x/2.95)]

The problem is You need to produce a buffer solution that has a pH of 5.21. You already have a solution that contains 10 mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74.

There are 2.95 mmol acetic acid.

Now I am supposed to plug the 2.95 into the acid part of the equation:
pH = pKa + log (base/acid)
and solve for x.

5.21 = 4.74 + log (x/2.95)

With "log" being where it is, I do not know how to isolate "x"

I see that you don't have an answer to your log question. I would suggest maybe checking out www.confusinghomework.com. They offer free online tutoring (full detailed steps and final answer) to your homework you need help with. I have used them for my calc 2 homework multiple times after not getting an answer on here.

5.21 = 4.74 + [log (x/2.950)]
0.47 = log (x/2.95)
10^0.47 = (x/2.95)
2.951209227 = (x/2.95)
x = 8.706

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