Question

(2^-n * 8^(2n+1) * 16^2n)/4^3n=1/16

Answer 1

ANSWER: -1

EXPLANATION: I have attached solution with the answer to help you to understand well.

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Answer 2

Answer:

-1 is the required value of n

Step-by-step explanation:

Explanation:

Given that, $\frac{[2^{-n}.8^{2n +1} . 16^{2n}]}{4^{3n}}$ = $\frac{1}{16}$

• According to the formula $a^m .a^n = a^{m+n}$ raising a number to a power equal to the sum of m + n is equivalent to multiplying it by the same number a by itself m times and then by the same number a multiplied by itself n times.

• And we also know that, the rule for an exponent $(a^m)^n = a^{mn}$

Step 1:

We have, $\frac{[2^{-n}.8^{2n +1} . 16^{2n}]}{4^{3n}}$ = $\frac{1}{16}$

This can be written as,

$\frac{[2^{-n}.(2^3)^{2n +1} . (2^4)^{2n}]}{4^{3n}}$ = $\frac{1}{16}$

Now by the exponent formula, $(a^m)^n = a^{mn}$

⇒ $\frac{[2^{-n}.(2)^{3(2n +1)} . (2)^{4(2n)}]}{4^{3n}}$ = $\frac{1}{2^4}$

⇒$\frac{[2^{-n}.(2)^{(6n +3)} . (2)^{(8n)}]}{4^{3n}}$ = $\frac{1}{2^4}$

Now, from the exponent formula $a^m . a^n = a^{m+n}$

⇒$\frac{[2^{-n+(6n+3)+8n}]}{2^{6n}}$ = $\frac{1}{2^4}$

⇒$2^{-n +6n + 3 +8n -6n} = 2^{-4}$

⇒$2^{7n +3 } = 2^{-4}$

In comparing both sides,

⇒7n + 3 = -4

⇒ 7n = -7

⇒n = $\frac{-7}{7}$ = -1

Final answer:

Hence, -1 is the required value of n.

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