Simplifying $\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$ 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: . 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?
 A: why not use base 16? You have
$$
\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}} = \\
\frac{16 \cdot  16^{x}+20  \cdot  16^{x}}{16^{x} 2^{-3}8^{2}} = \\
\frac{36}{8} =  \frac{9}{2}
$$
A: 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.
