Solving equations involving powers

Question

I have been solving an OS problem in which I came across this equation where I have to find the value of H as in % but I'm not able to solve it fully. This is not a homework, I'm stuck in between and been left maths from quite some time.

3*10^-8/0.8 = H(2*10^-8 + 10^-8) + (1-H) (10^-8 + 10^-3)

My Take so far -

3*10^-8/0.8 = H(3*10^-8) + (1-H) (10^-11)

3*10^-8/0.8 = H(3*10^-8) + 10^-11 - 10^-11*H

Now I can't able to solve any further.

What I tried more is I cancelled (3*10^-8) from both LHS and RHS and I got

1/0.8 = H + (1-H)*10-11 /// ( I took 10^-11 common from 10^-11 - 10^-11*H)

So what to do after 1/0.8 = H + (1-H)*10-11

Answer 1

So you have

$$\frac{3\cdot10^{-8}}{0.8} = 3\cdot10^{-8}H + (1-H)(10^{-8}+10^{-3})$$

which is

$$\frac{3}{0.8} = 3H + (1-H)(1+10^{5}) = 1+10^{5}-10^{5}H+2H$$

if you divide both sides by $10^{-8}$. So

$$H = \frac{\frac{3}{0.8} - 1 - 10^{5}}{2-10^{5}} = 0.9999924998499969999... \approx\mathbf{100\%}$$

Answer 2

Dropping the annoying $10^{-8}$,

$$\frac3{0.8} = H(2 + 1) + (1-H) (1 + 100000).$$

Without caring about lower terms, we have

$$1-H\approx\frac3{80000}$$ which gives you an order of magnitude, i.e. $H$ very close to $1$.

Exact computation is (using $H=1-(1-H)$)

$$\frac{15}4=3+(1-H)\,99998,$$

$$1-H=\frac3{4\cdot99998}.$$

(I prefer to evaluate $1-H$ as this is more accurate and gives you better understanding of the value.)