Solve inverse log + constant

Question

Is it possible to solve for b in the following?

a = log2(b + c)

For example:

3.907 = log2(b + 10)
b = 5

Answer 1

$$\begin{array}{rcl} a &=& \log_2(b+c) \\ 2^a &=& b+c \\ 2^a-c &=& b \\ b &=& 2^a-c \end{array}$$

Answer 2

Hint: $$y = \log_2(x) \Leftrightarrow 2^y=x $$