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If you simplify $3\cdot3^{2x}\cdot\frac{1}{4}\cdot4^{2x}\cdot25\cdot5^{3x}=6\cdot2^{2x}\cdot3^{2x}\cdot5^{2x}$, this can be written as $4^x\cdot5^x = \frac{a}{b}$. How do you express $\frac{a}{b}$ in simplest form?

I did the math and first turned the whole equation into an exponential format:
$3^{2x+1}\cdot4^{2x-1}\cdot5^{3x+2}=(2\cdot3)\cdot2^{2x}\cdot3^{2x}\cdot5^{2x}$
$3^{2x+1}\cdot4^{2x-1}\cdot5^{3x+2}=2^{2x+1}\cdot3^{2x+1}\cdot5^{2x}$
$3^{2x+1}\cdot4^{2x-1}\cdot\frac{5^{3x+2}}{5^{2x}}=2^{2x+1}\cdot3^{2x+1}\cdot$
$3^{2x+1}\cdot4^{2x-1}\cdot5^{x+2}=2^{2x+1}\cdot3^{2x+1}\cdot$
$4^{2x-1}\cdot5^{x+2}=2^{2x+1}\cdot\frac{3^{2x+1}}{3^{2x+1}}$
$4^{2x-1}\cdot5^{x+2}=2^{2x+1}$

Where do I go from here?

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    You are almost there: from the last line you can say $4^{x} \cdot 5^{x}=8/25$. All you need is to write $2^{2x+1}=4^{x} \cdot 2$ and then simplify so that only $4^x 5^x$ remains on the left. – 2017-01-12

2 Answers 2

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$4^{2x-1} = 4^x\cdot4^{x-1}, 5^{x+2} = 25\cdot5^x $, hence

$$4^{2x-1}\cdot5^{x+2}=2^{2x+1} \iff\\ 4^x\cdot4^{x-1}\cdot25\cdot5^x = 2^{2x+1}\iff\\ 4^x\cdot(2^2)^{x-1}\cdot25\cdot5^x = 2^{2x+1} $$

Can you take it from here?

Recall the exponent rule $(a^b)^c = a^{bc} $:

$$4^x\cdot(2^2)^{x-1}\cdot25\cdot5^x = 2^{2x+1}\iff\\ 4^x\cdot 2^{2x-2}\cdot25\cdot5^x = 2^{2x+1}\iff\\ 4^x\cdot5^x = \frac{2^{2x+1}}{25\cdot2^{2x-2}}\iff\\ 4^x\cdot5^x = \frac{2^{2x-2}\cdot2^3}{25\cdot2^{2x-2}}\iff\\ 4^x\cdot5^x = \frac{8}{25} $$

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    Ok the answer is 8/25 and I see where the denominator, 25, comes from but I'm still confused on how to get the numerator, 8. @RSerrao – 2017-01-12
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    @Axop51 I edited my answer. Is it clearer now? – 2017-01-12
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Hint: $4=2^2$, so $4^p=2^{2p}$.