# Find the values of variables [closed]

If $2^a+4^b+8^c=328$ Then how do I find the value of $\frac{a+2b+3c}{abc}$. Any help is appreciated.

## closed as off-topic by user91500, Claude Leibovici, user1551, Behrouz Maleki, tiredDec 19 '16 at 15:22

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• It seems like you forgot to write something (?). What are your attempts? – Eugenio Dec 19 '16 at 14:14
• 2a+2.2b+2.3c=328 – Shagun Yadav Dec 19 '16 at 14:28
• Now I cant find the way how to do it – Shagun Yadav Dec 19 '16 at 14:29
• Click on the "edit" button below your question and make the relevant changes to your question in order for me and others to understand. Take a quick look to MathJax to better write your questions meta.math.stackexchange.com/questions/5020/… – Eugenio Dec 19 '16 at 14:32
• Can you please solve the question – Shagun Yadav Dec 19 '16 at 14:33

Assuming your question means $$2^a+4^b+8^c=328$$ We proceed by dividing both sides by $8$. $$2^{a-3} + 2^{2b-3}+2^{3c-3} = 41$$ $$\Rightarrow 2^{a-3}+2^{2b-3}+8^{c-1} = 41$$ $c-1$ should be $1$. So, $c=2$. This gives, $$2^{a-3}+2^{2b-3} = 33$$ It can be easily checked that $a-3 = 0, 2b-3 = 5$. This gives $a=3, b=4$. The answer can now be easily obtained.
EDIT: Also possible that $a-3 = 5, 2b-3=0$.
• Think what are the powers of $8$. It can be $0,8,64\cdots$. A simple analysis for $8^0=1$ gives us no answers. The next case can be used. But, $64$ can't be possible as the sum of the other two quantities cannot be negative. – Rohan Dec 19 '16 at 14:39
Writing $328$ in base two, we get $$328=2^3+2^6+2^8$$ Even if $a,b,c\in\mathbb{N}$, we have three solutions: $$2^3+4^4+8^2=328\implies\frac{a+2b+3c}{abc}=\frac{17}{24}$$ $$2^6+4^4+8^1=328\implies\frac{a+2b+3c}{abc}=\frac{17}{24}$$ $$2^8+4^3+8^1=328\implies\frac{a+2b+3c}{abc}=\frac{17}{24}$$ Luckily, all three give the same value.