# How to solve this simulatenous equations which invovles exponents?

I am trying to find $x-y$ when: \begin{cases} 2^{x} - 2^{y} = 1 \\ 4^{x} - 4^{y} = \frac{5}{3}\\ \end{cases}

Simultaneously. From the second equation I factorized to get $$(2^x + 2^y)(2^x - 2^y) = \frac53$$

and from the first equation, since $(2^x - 2^y) = 1$, I dedeuced that $$(2^x + 2^y)\cdot 1 = \frac53 \to (2^x + 2^y) = \frac53$$

Now I am a little stuck on how to proceed. Any ideas how to carry on with explanations please?

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I have edited to add in what you really want - always wise to ask the whole question, because then you will get better help. I've also updated my answer. – Mark Bennet Mar 9 '14 at 19:12

Try adding the equations $2^x-2^y=1$ and $2^x+2^y=\frac 53$

To go further per your comment:

This gives $2\cdot 2^x=\frac 83$

Subtracting the equations gives $2\cdot 2^y=\frac 23$

Now recall that $\frac {2^a}{2^b}=2^{a-b}$ and divide in an appropriate way to make further progress.

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Thanks. How would I determine what $x - y$ is? Because the original homework question was to determine the value of $x-y$ so I thought it would be a good idea to find the values of x and y but that will not help me determine the value of $x - y$ because I don't know anything about log and I can't use a calculator. – Aspiring Mathlete Mar 9 '14 at 19:05

Well, now it's really simple because you have \begin{cases} 2^x - 2^y = 1 \\ 2^x + 2^y = \frac53 \\ \end{cases} so $$2^x - 2^y + 2^x - 2^y = 2\cdot 2^x = 1+\frac53 = \frac83 \Rightarrow 2^x = \frac43 \Rightarrow x = log_2{\frac43}$$ Finally, since $2^x = \frac43$, $$2^y = \frac53 - \frac43 = \frac13 \Rightarrow y = log_2{\frac13}$$ The solution is $\left(log_2{\frac43} , log_2{\frac13}\right)$.

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