Formula to find the exponent with a specific answer Here is the equation
$$\frac{100}{0.003(2^x)}=?$$
The answer needed is in the range $(1,2)$.
How can I compute the exponent so this equation will get the desired answer? Here is a sample.
$$\frac{100}{0.003(2^{15})} = 1.01725260417$$
Please help. Thanks!
 A: I am going to first define
$$f(x)=\frac{100}{0.003(2^x)}$$
So basically we need
$$1<f(x)<2$$
which is
$$1<\frac{100}{0.003(2^x)}<2$$
If you are strong with inequalities, you can use perform operations in one go. Otherwise, I suggest you splitting it up into
$$\frac{100}{0.003(2^x)}>1\quad\text{and}\quad
\frac{100}{0.003(2^x)}<2$$
Since all the numbers are positive and both the exponential and logarithmic functions are strictly increasing you can easily move things around to get, for the left hand side,
\begin{align}
\frac{100}{0.003(2^x)}&>1\\
\frac{100}{0.003}&>2^x\\
x<\log_2{\left(\frac{100}{0.003}\right)}\approx15.0247
\end{align}
and, for the right hand side,
\begin{align}
\frac{100}{0.003(2^x)}&<2\\
\frac{50}{0.003}&<2^x\\
x>\log_2{\left(\frac{50}{0.003}\right)}\approx14.0247
\end{align}
Hence,
$$\log_2{\left(\frac{50}{0.003}\right)}<x<\log_2{\left(\frac{100}{0.003}\right)}$$
or
$$14.0247\lesssim x\lesssim15.0247$$
A: $$1 < \frac{100}{0.003(2^x)} < 2$$
$$\ln 1 < \ln\left(\frac{100}{0.003(2^x)}\right) < \ln 2$$
$$ 0 < \ln(100) - \ln(0.003) - x\ln2 < \ln 2$$
$$ 0 < 4.6051 - (-5.8091) - 0.6931 \times x < 0.6931$$
$$ 0 < 10.4143 - 0.6931 \times x < 0.6931$$
$$ 0 < 15.0247 - x < 1$$
$$ 14.0247 < x < 15.0247$$
