Question

please solve this question​

Answer 1

Answer:

Here is the solution

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Answer 2

Question

Solve ⇒ $\sf \dfrac{36\times (-6)^{2}\times 3^{6} }{12^{3} \times 3^{5} }$

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Answer

In the given question we must make all the numbers in the prime form or simplest form. So let's do it.

⇒ $\sf \dfrac{36\times (-6)^{2}\times 3^{6} }{12^{3} \times 3^{5} }$

In this 36 can be written as 2×2×3×3. But we need to make it into exponential value so our number 36 would be written as ⇒ 2²×3². So we will write 36 as 2²×3² in the problem.

⇒ $\sf \dfrac{36\times (-6)^{2}\times 3^{6} }{12^{3} \times 3^{5} }$

⇒ $\sf \dfrac{2^{2}\times 3^{2} \times (-6)^{2}\times 3^{6} }{12^{3} \times 3^{5} }$

Now 12 can be written as 2×2×3. But we need to make it into exponential value so our number 13 would be written as ⇒ 2²×3. So we will convert 12 to 2²×3.

⇒ $\sf \dfrac{2^{2}\times 3^{2} \times (-6)^{2}\times 3^{6} }{12^{3} \times 3^{5} }$

⇒ $\sf \dfrac{2^{2}\times 3^{2} \times (-6)^{2}\times 3^{6} }{(2^{2}\times 3) ^{3} \times 3^{5} }$

Now we will make (-6)² as it's original value and then simplify it. (-6)² = (-6)×(-6) = 36. In this 36 can be written as 2×2×3×3. But we need to make it into exponential value so our number 36 would be written as ⇒ 2²×3². So we will write (-6)² as 2²×3² in the problem.

⇒ $\sf \dfrac{2^{2}\times 3^{2} \times (-6)^{2}\times 3^{6} }{(2^{2}\times 3) ^{3} \times 3^{5} }$

⇒ $\sf \dfrac{2^{2}\times 3^{2} \times 2^{2}\times 3^{2} \times 3^{6} }{(2^{2}\times 3) ^{3} \times 3^{5} }$

Now we will apply this law of exponent to remove brackets ⇒ $(a^{m})^{n}=a^{m\times n}$

⇒ $\sf \dfrac{2^{2}\times 3^{2} \times 2^{2}\times 3^{2} \times 3^{6} }{(2^{2}\times 3) ^{3} \times 3^{5} }$

⇒ $\sf \dfrac{2^{2}\times 3^{2} \times 2^{2}\times 3^{2} \times 3^{6} }{2^{2\times 3 }\times 3^{1\times 3} \times 3^{5} }$

⇒ $\sf \dfrac{2^{2}\times 3^{2} \times 2^{2}\times 3^{2} \times 3^{6} }{2^{6}\times 3^{3} \times 3^{5} }$

Now we will combine like bases together and rearrange the numbers.

⇒ $\sf \dfrac{2^{2}\times 3^{2} \times 2^{2}\times 3^{2} \times 3^{6} }{2^{6}\times 3^{3} \times 3^{5} }$

⇒ $\sf \dfrac{2^{2}\times 2^{2}\times 3^{2} \times3^{2} \times 3^{6} }{2^{6}\times 3^{3} \times 3^{5} }$

Next, we will apply this law of exponent ⇒ $a^{m}\times a^{n} = a^{m+n}$

⇒ $\sf \dfrac{2^{2}\times 2^{2}\times 3^{2} \times3^{2} \times 3^{6} }{2^{6}\times 3^{3} \times 3^{5} }$

⇒ $\sf \dfrac{2^{2+2}\times 3^{2+2} \times 3^{6}}{2^{6}\times 3^{3+5}}$

⇒ $\sf \dfrac{2^{4}\times 3^{4+6}}{2^{6}\times 3^{8}}$

⇒ $\sf \dfrac{2^{4}\times 3^{10}}{2^{6}\times 3^{8}}$

Now we must apply this law of exponent ⇒ $a^{m}\div a^{n} = a^{m-n}$

⇒ $\sf \dfrac{2^{4}\times 3^{10}}{2^{6}\times 3^{8}}$

⇒ $\sf 2^{4-6}\times 3^{10-8}$

⇒ $\sf 2^{-2}\times 3^{2}$

Now we will apply negative power law ⇒ $a^{-m}=\frac{1}{a^{m}}$

⇒ $\sf 2^{-2}\times 3^{2}$

⇒ $\sf \dfrac{1}{2^{2}} \times 3^{2}$

At last, we will give the exponents their actual value and multiply.

⇒ $\sf \dfrac{1}{2^{2}} \times 3^{2}$

⇒ $\sf \dfrac{1}{(2\times 2) } \times (3\times 3)$

⇒ $\sf \dfrac{1}{4 } \times 9$

⇒ $\sf \dfrac{9}{4}$

$\bf \therefore \dfrac{36\times (-6)^{2}\times 3^{6} }{12^{3} \times 3^{5} } =\dfrac{9}{4}$

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