Solveeit Logo

Question

Question: $\sqrt{\log_2 3 \log_2 12 \log_2 48 \log_2 192 - \log_2 12 \log_2 48 + 10}$...

log23log212log248log2192log212log248+10\sqrt{\log_2 3 \log_2 12 \log_2 48 \log_2 192 - \log_2 12 \log_2 48 + 10}

Answer

None of the above

Explanation

Solution

To simplify the given expression, let x=log23x = \log_2 3.

We can express the other logarithmic terms in terms of xx:

log212=log2(223)=log2(22)+log23=2+x\log_2 12 = \log_2 (2^2 \cdot 3) = \log_2 (2^2) + \log_2 3 = 2 + x

log248=log2(243)=log2(24)+log23=4+x\log_2 48 = \log_2 (2^4 \cdot 3) = \log_2 (2^4) + \log_2 3 = 4 + x

log2192=log2(263)=log2(26)+log23=6+x\log_2 192 = \log_2 (2^6 \cdot 3) = \log_2 (2^6) + \log_2 3 = 6 + x

Let the expression inside the square root be EE.

E=log23log212log248log2192log212log248+10E = \log_2 3 \log_2 12 \log_2 48 \log_2 192 - \log_2 12 \log_2 48 + 10

Substitute the terms in terms of xx:

E=x(2+x)(4+x)(6+x)(2+x)(4+x)+10E = x (2+x) (4+x) (6+x) - (2+x) (4+x) + 10

Let's rearrange the product of the first four terms:

x(2+x)(4+x)(6+x)=[x(6+x)][(2+x)(4+x)]x (2+x) (4+x) (6+x) = [x(6+x)] [(2+x)(4+x)]

=(x2+6x)(x2+4x+2x+8)= (x^2 + 6x) (x^2 + 4x + 2x + 8)

=(x2+6x)(x2+6x+8)= (x^2 + 6x) (x^2 + 6x + 8)

Let K=x2+6xK = x^2 + 6x.

Then the product becomes K(K+8)K(K+8).

The second term in the original expression is (2+x)(4+x)=x2+6x+8=K+8(2+x)(4+x) = x^2 + 6x + 8 = K+8.

Now substitute these into the expression for EE:

E=K(K+8)(K+8)+10E = K(K+8) - (K+8) + 10

Factor out (K+8)(K+8) from the first two terms:

E=(K+8)(K1)+10E = (K+8)(K-1) + 10

Expand the product:

E=K2K+8K8+10E = K^2 - K + 8K - 8 + 10

E=K2+7K+2E = K^2 + 7K + 2

Substitute K=x2+6xK = x^2 + 6x back into the expression for EE:

E=(x2+6x)2+7(x2+6x)+2E = (x^2 + 6x)^2 + 7(x^2 + 6x) + 2

E=(x4+12x3+36x2)+(7x2+42x)+2E = (x^4 + 12x^3 + 36x^2) + (7x^2 + 42x) + 2

E=x4+12x3+43x2+42x+2E = x^4 + 12x^3 + 43x^2 + 42x + 2

The expression inside the square root is x4+12x3+43x2+42x+2x^4 + 12x^3 + 43x^2 + 42x + 2, where x=log23x = \log_2 3.

Since x=log23x = \log_2 3 is an irrational number, this expression does not simplify to a simple integer. This suggests a potential typo in the question, as such problems in competitive exams usually simplify to a rational number, often an integer.

However, if we are forced to provide a solution based on the given question, the simplified form of the expression is K2+7K+2K^2 + 7K + 2, where K=(log23)2+6log23K = (\log_2 3)^2 + 6 \log_2 3.

Given that the similar question provided has a numerical answer of 10, despite its explanation indicating the expression doesn't simplify to a constant, it points towards a common pattern where such expressions are designed to simplify.

If the expression was designed to simplify, it would typically be a perfect square of an integer or a constant.

The expression K2+7K+2K^2+7K+2 is not a perfect square of a linear term in KK, nor is it a constant.

Without further clarification or correction to the question, a numerical answer cannot be obtained. Assuming the question is correctly stated and it is expected to simplify to an integer, there might be a very subtle identity or a specific property of log23\log_2 3 that makes it simplify, which is not immediately apparent. However, typically, such problems simplify to a constant value independent of xx.

If the question were slightly different, for example, if the expression inside the square root was (K+1)2=K2+2K+1(K+1)^2 = K^2+2K+1 or (K+2)2=K2+4K+4(K+2)^2 = K^2+4K+4, or (K+3)2=K2+6K+9(K+3)^2 = K^2+6K+9, it would simplify. But we have K2+7K+2K^2+7K+2.

In the absence of any choices or further context suggesting a specific value, the expression cannot be simplified to a single numerical value.