Compute this equation of logarithms under a squareroot

I'll just focus to the harsh square root over there. Note that inside of the logarithms can be factored. Manipulating stuff and we get $\begin{aligned} \sqrt{\log_{2}(3)\log_{2}(12)\log_{2}(48)\log_{2}(192)+16} &= \sqrt{\log_{2}(3)\log_{2}(3\cdot 2^2)\log_{2}(3\cdot 2^4)\log_{2}(3\cdot 2^6)+16} \\ &= \sqrt{\log_{2}(3)[\log_{2}(3) + 2][\log_{2}(3) + 4][\log_{2}(3) + 6]+16} \end{aligned}$ Let $y = \log_{2}(3) + 3$ . Then the expression becomes $\begin{aligned} \sqrt{(y - 3)(y - 1)(y + 1)(y + 3)+16} &= \sqrt{(y^2 - 1)(y^2 - 9)+16} \\ &= \sqrt{(y^4 - 10 y^2 + 9)+16} \\ &= \sqrt{y^4 - 10 y^2 + 25} = \sqrt{(y^2 - 5)^2} = y^2 - 5 \end{aligned}$ The value requested now is just $\begin{aligned} &\sqrt{\log_{2}(3)\log_{2}(12)\log_{2}(48)\log_{2}(192)+16} - \log_{2}(12)\log_{2}(48) + 10 \\ =&(\log_{2}(3) + 3)^2 - 5 - (\log_{2}(3) + 2)(\log_{2}(3) + 4) + 10 \\ =&(\log_{2}^{2}(3) + 9 + 6\log_{2}(3)) - 5 - (\log_{2}^{2}(3) + 6\log_{2}(3) + 8) + 10 = \boxed{6} \end{aligned}$

$\begin{aligned} x & = \sqrt{\log_2(3)\log_2(12)\log_2(48)\log_2(192)+16} - \log_2(12)\log_2(48) + 10 \\ & = \sqrt{\log_2(3)\log_2(4\times 3)\log_2(16\times 3)\log_2(64\times 3)+16} - \log_2(4\times 3)\log_2(16\times 3) + 10 \\ & = \sqrt{\log_2 3(2+\log_23)(4+\log_2 3)(6+\log_2 3)+16} - (2+\log_23)(4+\log_2 3) + 10 & \small \color{#3D99F6} \text{Let }\log_2 3 = a \\ & = \sqrt{a(2+a)(4+a)(6+a)+16} - (2+a)(4+a) + 10 \\ & = \sqrt{a^4+12a^3+44a^2 +48a+16} - (a^2+6a+8) + 10 \\ & = \sqrt{(a^2+6a+8)^2-8(a^2+6a+8)+16} - (a^2+6a+8) + 10 \\ & = (a^2+6a+8) - 4 - (a^2+6a+8) + 10 \\ & = \boxed 6 \end{aligned}$

Log in the solution below refers to Log to the base 2.

Log(12) = Log(3*4) = Log(3) + Log(4) = Log(3) + 2;

Similarly Log(48) = Log(3) + Log(16) = Log(3) + 4; Log(192) = Log(3) + 6.

So the expression under the root is Log(3)(Log(3) +2)(Log(3)+4)(Log(3)+6) + 16.

Let us take Log(3) as equal to some var Y , the expression under the root can be written as as Y*(Y+2)(Y+4)(Y+6) + 16;

Reorder the product as Y(Y+6)*(Y+4)(Y+2) + 16 or (Y^2 + 6Y)(Y^2+6Y+8) + 16 , substitute Y^2 + 6Y as t ,so the expression becomes t^2 + 8t + 16 or just (t+4)^2.

So the expression under the root reduces to t+4 where t is...;

Now let us look at the expression outside the root -(log3 + log4)(log3 +log16) + 10 = -(log(3) + 2)(log(3) + 4) = -(X+2)(X+4) + 10 = - ( X^2 + 6X + 8) + 10 or -(t +8) +10 or 2 - t.

The sum of the entire expression reduces to t+4 + 2 - t = 6. So X = 6.