# Intricate exponential equation

This is the question: $$\frac{(2^{3n+4})(8^{2n})(4^{n+1})}{(2^{n+5})(4^{8+n})} = 2$$ I've tried several times but I can't get the answer by working out.I know $n =2$, can someone please give me some guidance? Usually I turn all the bases to 2, and then work with the powers, but I probaby make the same mistake every time, unfortunately I don't know what that is. Thank you in advance.

EDIT

This is what I simplified it to in the beginning of every attempt.

$$\frac{(2^{3n} *16)(2^{6n})(2^{2n}*4)}{(2^{n}*32)(2^{n}*2^{16})} = 2$$

Therefore

$$\frac{64(2^{3n+6n+2n})}{(2^{16}*32)(2^{2n})} = 2$$
I simplified further: $$\frac{2^{11n}}{32768(2^{2n})} =2$$
$$2^{11n} = (2^{2n+1})*32768 \\$$ $$\frac{2^{11n}}{2^{2n+1}} = 32768$$

$$\frac{2^{11n}}{2*2^{2n}} = 32768$$

And this is the furthest I get, what do i do now?

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If you show us what you do, we'll show you the mistake. –  Gerry Myerson Mar 9 at 4:46
@GerryMyerson I just did, thank you in advance! –  Samir Chahine Mar 9 at 5:04
$4^{8+n}\ne2^n2^{16}$. –  Gerry Myerson Mar 9 at 8:50
I didn't even realize! Thank you, makes much more sense! –  Samir Chahine Mar 9 at 10:08

We have \begin{align} \frac{(2^{3n+4})(8^{2n})(4^{n+1})}{(2^{n+5})(4^{8+n})} &= \frac{(2^{3n+4})(2^{3})^{2n}(2^2)^{n+1}}{(2^{n+5})(2^2)^{8+n}} \\ &=\frac{2^{3n+4+6n+2n+2}}{2^{n+5+16+2n}} \\ &=\frac{2^{11n+6}}{2^{3n+21}} \\ &=2^{(11n+6)-(3n+21)} \\ &=2^{8n-15} \end{align}

Also from the original problem, \begin{align} \frac{(2^{3n+4})(8^{2n})(4^{n+1})}{(2^{n+5})(4^{8+n})} &= 2 \end{align}

Therefore, \begin{align} 2^{8n-15} &= 2^1 \end{align}

Drop the base $2$ on both sides, and we will get \begin{align} 8n-15 = 1 \end{align} Thus, $n=2$.

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That's much simpler than what I tried to do! Thank you! Much appreciated :) –  Samir Chahine Mar 9 at 5:09
No problem! :-D –  le gâteau au fromage Mar 9 at 5:10

HINT:

Use $$a^{mn}=(a^m)^n$$ and $$a^m\cdot a^n=a^{m+n}\text{ and }\frac{a^m}{a^n}=a^{m-n}$$

For example, $\displaystyle4^{8+n}=(2^2)^{(8+n)}=2^{2(8+n)}$

Finally $a^x=a^y\implies x=y$ if real $a\ne\pm1$

Reference : Exponent Combination Laws

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Also, notice that 4 and 8 are both powers of 2. –  Gerry Myerson Mar 9 at 4:45
@GerryMyerson, that's the basis, else things will be really complex –  lab bhattacharjee Mar 9 at 4:52

$$\frac{2^{11n}}{2*2^{2n}} = 32768$$

And this is the furthest I get, what do I do now?

Multiply both sides by 2. Then you get $$\frac{2^{11n}}{2^{2n}} = 65536.$$ $$\implies 2^{11n-2n}=2^{9n}=65536.$$ $$\therefore n=\frac{log_2(65536)}{9}=\frac{16}{9}.$$

EDIT: This is wrong, and there should be a $3n$ instead of a $2n$ in the denominator to start with, yielding $n=2$ because $4^{8+n}=4^{16+2n}$, not $4^{16+n}$ as in the question.

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$n = 2$ though, I made a mistake, but I know what it is now, thank you! –  Samir Chahine Mar 9 at 5:14
The additions in your comment, Mathster, should be multiplications. –  Gerry Myerson Mar 9 at 8:51
@GerryMyerson You're right. I deleted my comment. By the way, by far the easiest way is to start off by taking $log_2$ of both sides. Then you have $3n+4+3(2n)+2(n+1)-[(n+5)+2(8+n)]=1.$ Solving for $n$ to yield the result of $n=2$ is then [even more] trivial. –  WChang Mar 9 at 19:43