If $2^x=7$ and $2^y = 14$, then $x-y=$ ?

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$$2^x= 7, 2^y = 14$$ $$\frac{2^x}{2^y} = \frac{7}{14}$$ $$2^{x-y} = \frac{1}{2}$$ $$2^{x-y} = 2^{-1}$$ $$x-y = \boxed{-1}$$

Use the calculator to manually calculate $x$ and $y$. You can do this by changing it into logarithmic form, such as

$$2^x=7$$ $$\log_22^x = \log_27$$ $$x = \log_27$$

After obtaining $x$ and $y$ with your calculator, just subtract them.

Technically, you can manually get the answer if you know logarithm rules:

$$2^y=14$$ $$\log_22^y = \log_214$$ $$y=\log_214$$ $$x-y = \log_27 -\log_214$$ $$= \log_2\left(\frac{7}{14}\right)$$ $$= \log_2\left(\frac{1}{2}\right)$$ $$= \log_2(2^{-1})$$ $$= -1\log_22$$ $$= \boxed{-1}$$