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Simplify the following: $\left ( 8^{4+2a} \right )\left ( 16^{a-1} \right )\div 4^{3a+2}$
I don't know how to simplify this expression.

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    $\begingroup$ $8^{4+2a}=\Big(2^3\Big)^{4+2a}$ $=2^{3(4+2a)}$ $=2^{12+6a}$. And just as $8$ is $2^3$, so $16$ is $2^4$ and $4$ is $2^2$. ${}\qquad{}$ $\endgroup$ Feb 14, 2014 at 1:26

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Hint: Use that $$4=2^2\qquad 8=2^3\qquad 16=2^4$$ as well as these rules about exponentiation: $$(a^b)^c=a^{(bc)}\qquad a^b\div a^c=a^{b-c}$$

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  • $\begingroup$ Thank you very much :) This hint helped me a lot. $\endgroup$ Feb 14, 2014 at 1:42
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Note that: $$8^{4+2a}=\left(2^3\right)^{4+2a}=2^{3(4+2a)}$$ $$16^{a-1}=\left(2^4\right)^{a-1}=2^{4(a-1)}$$ $$4^{3a+2}=\left(2^2\right)^{3a+2}=2^{2(3a+2)}$$

Since they are all raised to the same base $(2)$ you can go ahead and apply the relevant exponential laws to simplify your expression. Namely $a^na^m=a^{n+m}$ to simplify the numerator and $\frac{a^n}{a^m}=a^{n-m}$ to further simplify the remaining expression.

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  • $\begingroup$ No problem, glad I could help! $\endgroup$
    – Zhoe
    Feb 14, 2014 at 1:42
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$\log_2(n)\, =\, (4\!+\!2a)\,3 + (a\!-\!1)\,4 - (3a\!+\!2)\,2\, =\, 4\!+\!4a\, =\, 4(1\!+\!a),\ $ so $\ n = 2^{4(1+a)} = 16^{1+a}$

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Hint: Convert each number to same base. What ever algorithm you do to a numbers those must be in same base. so in this problem convert to base 2.then do your usual calculation on indices.

Then game is over.

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