# 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. $$\frac{100}{0.003(2^{15})} = 1.01725260417$$

• Not following. Are you asking "for which $x$ is $1< \frac {100}{.003\times 2^x}<2$?" – lulu Apr 25 '18 at 9:51
• It's possible to solve for $x$ in this case using relatively simple algebra. Have you tried that? – Matti P. Apr 25 '18 at 9:53
• @lulu x is the exponent needed to have the answer of the equation is >1 but <2. – Mix Austria Apr 25 '18 at 9:54
• @MattiP. im sorry I'm not really good at math. I just need this in a script that I'm programming :( – Mix Austria Apr 25 '18 at 9:57

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

$$14.0247\lesssim x\lesssim15.0247$$

$$1 < \frac{100}{0.003(2^x)} < 2$$ $$\ln 1 < \ln\left(\frac{100}{0.003(2^x)}\right) < \ln 2$$ $$0 < \ln(100) - \ln(0.003) - x\ln2 < \ln 2$$ $$0 < 4.6051 - (-5.8091) - 0.6931 \times x < 0.6931$$ $$0 < 10.4143 - 0.6931 \times x < 0.6931$$ $$0 < 15.0247 - x < 1$$ $$14.0247 < x < 15.0247$$