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I'm struggling trying to simplify $$\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$$

Here's my attempt:

All of these coefficients can be converted to base $2$ exponentials.

$$\frac{2^{4^{(x+1)}}+20^\ (2^{2^{(2x)}})}{2^{x-3}2^{3^{(x+2)}}}$$

Doing some algebra:

$$\frac{2^{4x+4}+20^\ (2^{4x})}{2^{x-3}2^{3x+6}}$$

The denominator can be simplified.

$$\frac{2^{4x+4}+20^\ (2^{4x})}{2^{4x+3}}$$

Now, here's where I think I've made a mistake, but I'm gonna try to simplify it further by splitting this up into separate fractions.

$$\frac{2^{4x+4}}{2^{4x+3}} + \frac{5 \bullet 2^2 \ 2^{4x}}{2^{4x+3}}$$

$$2\ \frac{2^{4x+3}}{2^{4x+3}} + \frac{5\ \bullet 2^{4x+2}}{2^{4x+3}}$$

$$2 + 5\ \bullet 1/2\ \frac{2^{4x+2}}{2^{4x+2}}$$

$$2 + 5/2$$

$$4/2 + 5/2$$

$$9/2$$

Whelp, I just solved it properly while putting it in as a question. However, I still have two queries:

  • Symbolab's answer is this: enter image description here. Is there logic behind this answer or is outright not simplified as well?

  • Does this algorithm to solving (changing all to the right base and trying to do algebra do them apply to ones even if they aren't connected by whole number bases? Like if I were trying to do one with base $5$ and base $7$ exponentials, could I solve it using logarithm laws just the same?

sangstar
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  • your solution seems correct. – Arnaldo Jul 18 '17 at 20:23
  • I'd approach it by prime factorizing the top and bottom which is what you've done. However, spotting that there is a factor of $16^x$ on both sides as @Andreas has pointed out is quite a nice shortcut that'll save a few lines of working. If you find that the top and bottom don't share prime factors (or bases, if you prefer), then there isn't much you can do to simplify it in most cases. – Shuri2060 Jul 18 '17 at 21:29

2 Answers2

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why not use base 16? You have

$$ \frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}} = \\ \frac{16 \cdot 16^{x}+20 \cdot 16^{x}}{16^{x} 2^{-3}8^{2}} = \\ \frac{36}{8} = \frac{9}{2} $$

Andreas
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1

You just need to get more comfortable converting bases. Then any way you want to do it will work:

$\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$

$=\frac {2^{4(x+1)} + 5*4*2^{2*2x}}{2^{x-3}2^{3(x+2)}}$

$=\frac {2^{4x + 4} + 5*2^22^{4x}}{2^{(x-3)+3x + 6}}$

$=\frac {2^{4x+4} + 5*2^{4x+2}}{2^{4x + 3}}$

$= \frac {2^{4x+2}(2^2 + 5)}{2^{4x + 3}}$

$=9*\frac {2^{4x+2}}{2^{4x + 3}}$

$= 9*\frac 1{2^{(4x + 3)-(4x+2)}}$

$= 9*\frac 12$

"Does this algorithm to solving (changing all to the right base and trying to do algebra do them apply to ones even if they aren't connected by whole number bases? Like if I were trying to do one with base 5 and base 7 exponentials, could I solve it using logarithm laws just the same?"

Yes, but it might not be easy. And without calculators it might not be possible.

Ex: $\frac {7^x + 3^{2x}}{2^{3x + 1}} =$

$\frac {2^{x\log_2 7} + 2^{2x\log_2 3}}{2^{3x+1}}=$

$2^{x\log_2 7 + 2x \log_2 3 - 3x - 1} =$

$2^{x(\log_2 (63) -3) - 1}$.

But is that really "simpler"?

Post-script.... Actually I guess it is.

If you need to solve $\frac {7^x + 3^{2x}}{2^{3x + 1}} = 43$ we get

$2^{x(\log_2 (63) -3) - 1} = 43$ so

$x(\log_2 (63) - 3) - 1 = \log_2 43$

$x = \frac {\log_2 43 + 1}{\log_2(63) - 3}$

$\approx 2.1584348532293047695329252628754$.

We didn't have to do this base $2$. We could have used any base.

fleablood
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