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(2^(4-1)^2)/(2^(3-2))

This topic has 2 expert replies and 0 member replies

(2^(4-1)^2)/(2^(3-2))

Post Fri Nov 10, 2017 6:12 am
$$\frac{2^{\left(4-1\right)^2}}{2^{\left(3-2\right)}}=0$$

A. 2^8
B. 2^7
C. 2^6
D. 2^5
E. 2^4

The OA is A .

I don't know how to do the calculus here. Experts, can you help me?

Thanks.

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Post Fri Nov 10, 2017 7:28 am
Hello Vincen.

Let's do the calculus step by step.
$$1-\,\, \left(4-1\right)^2=\left(3\right)^2=9.$$ So, $$2^{\left(4-2\right)^2}=2^9.$$ Now, $$2-\,\, \ 2^{\left(3-2\right)}=2^1.$$ Finally, $$3-\,\, \ \frac{2^{\left(4-1\right)^2}}{2^{\left(3-2\right)}}=\frac{2^9}{2^1}=2^{\left(9-1\right)}=2^{8.}$$ So, the correct answer is A .

I hope this can help you.

I'm available if you'd like a follow up.

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Post Fri Nov 10, 2017 8:30 am
Vincen wrote:
$$\frac{2^{\left(4-1\right)^2}}{2^{\left(3-2\right)}}=0$$

A. 2^8
B. 2^7
C. 2^6
D. 2^5
E. 2^4

The OA is A .

I don't know how to do the calculus here. Experts, can you help me?

Thanks.
I have a feeling that you didn't mean to write that the expression $$\frac{2^{\left(4-1\right)^2}}{2^{\left(3-2\right)}}$$ equals ZERO.
This would be impossible.

Cheers,
Brent

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