My math skills are horrible. How do I solve for shutter speed time for a pinhole camera using the equation,
$$EV = 3.32 \log_{10} {N^2\over t}$$
\$N\$ = f-stop, in my case, f/195; \$EV\$ can be obtained from my light meter.
I would like to know my exposure time based on lighting conditions. I wish the EV calculation was linear!
I presume you are using the \$\log_{10}\$ from the question, Math of EV numbers. I wouldn't do that.
Instead, using the logarithm base-2 equation:
$$\begin{align} EV &= \log_2 {N^2\over t} \end{align}$$
Exponentiating both sides base 2 (in order to get the \$t\$ out of the base-2 logarithm):
$$\begin{align} 2^{EV} &= {N^2\over t} \\ \end{align}$$
Multiplying both sides by \$t\$ and dividing by \$2^{EV}\$ gives: $$\begin{align} t &= {N^2 \over 2^{EV}} \end{align}$$
Let's double check with our intuition. If the exposure value increases by 1, that means there's twice as much light in the scene, and the right hand side of the equation is smaller by a factor of 2. That means, if your f-number is constant, the shutter speed needs to decrease by a factor of 2 to get the same exposure.
And thinking in terms of aperture, for a constant \$EV\$, if your aperture changed from, say, 1.4 to 2.0 (one aperture stop, or one f-number, less light), then the \$N^2\$ in the right hand side means you need to double your shutter speed to get the same exposure.
I would like to know my exposure time based on lighting conditions. I wish the EV calculation was linear!
Well, the calculation is linear, if you think in terms of number of stops (i.e., count powers of two). While it's not difficult to learn to do so, unless you deal with powers of 2 often (like you would in say, computer science, computer engineering, or programming), it's hard to maintain that facility of working with numbers as if they're powers of 2.
For your case of f/195, it's roughly halfway between 128 and 256, which correspond to \$2^7\$ and \$2^8\$. So using \$\log_2 195 \approx 7.5\$ (it's actually 7.61), then \$\log_2 {N^2} = 2\log_2 {N} \approx 15 \$ (using the rules of logarithms).
Then, if you're shooting outside on a bright sunny day, your measured EV will probably be 15–16 or so. So
$$\begin{align} \log_2{t} &= {\sim}{15}\text{ (from EV) } - 15 \text{ (from }\log_2 N^2) \\ &\approx 0 \end{align}$$
So your exposure time should be about \$2^0 = 1\$ second. And for every EV your meter reads below about 15, you should double that time.
Comment (film): Because you are shooting in low light (that is, pinhole photography), if you are capturing onto film, you need to be aware of reciprocity failure (see What is "reciprocity failure"?). Bottom line, you might need to factor in a +25–50% increase in exposure time on top of what you calculate, because the low number of photons hitting the film grains (or in this case, the probably relatively long time between successive photons hitting the same film grain) causes the film to be less sensitive than nominal calculations would suggest. Each film type/brand has different reciprocity sensitivity characteristics; you'll need to consult their instructions for your particular film.
