If log64x2+log8y+3log512(yz)=4,wherex, yandzarepositiverealnumbers,thentheminimumpossiblevalueof(x+y+z)$ is
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Official Correct Answer: D. First, the equation given is:
$$
\log_{64} x^2 + \log_8 \sqrt{y} + 3\log_{512} \left( \sqrt{yz} \right) = 4
$$
Hmm, logarithms with different bases. logarithms can be converted to a common base using the change of base formula. Maybe that will help simplify things.
Let me recall the change of base formula:
$$
\log_b a = \frac{\log_c a}{\log_c b}
$$
So, I can convert all the logarithms to the same base, maybe base 2 since 64, 8, and 512 are all powers of 2. 64 is 2^6, 8 is 2^3, and 512 is 2^9. That should make things easier.
Let me rewrite each term:
1. First term: $$\log_{64} x^2$$
Using change of base to base 2:
$$
\log_{64} x^2 = \frac{\log_2 x^2}{\log_2 64} = \frac{2 \log_2 x}{6} = \frac{\log_2 x}{3}
$$
2. Second term: $$\log_8 \sqrt{y}$$
Again, change of base to base 2:
$$
\log_8 \sqrt{y} = \frac{\log_2 \sqrt{y}}{\log_2 8} = \frac{\frac{1}{2} \log_2 y}{3} = \frac{\log_2 y}{6}
$$
3. Third term: $$3\log_{512} \left( \sqrt{yz} \right)$$
First, apply the change of base:
$$
\log_{512} \sqrt{yz} = \frac{\log_2 \sqrt{yz}}{\log_2 512} = \frac{\frac{1}{2} (\log_2 y + \log_2 z)}{9} = \frac{\log_2 y + \log_2 z}{18}
$$
So, multiplying by 3:
$$
3 \times \frac{\log_2 y + \log_2 z}{18} = \frac{\log_2 y + \log_2 z}{6}
$$
Now, putting all the terms back into the original equation:
$$
\frac{\log_2 x}{3} + \frac{\log_2 y}{6} + \frac{\log_2 y + \log_2 z}{6} = 4
$$
Let me combine the terms:
First, let's find a common denominator, which is 6:
$$
\frac{2 \log_2 x}{6} + \frac{\log_2 y}{6} + \frac{\log_2 y + \log_2 z}{6} = 4
$$
Combine all the numerators:
$$
2 \log_2 x + \log_2 y + \log_2 y + \log_2 z = 24
$$
Simplify:
$$
2 \log_2 x + 2 \log_2 y + \log_2 z = 24
$$
Factoring out the coefficients:
$$
\log_2 x^2 + \log_2 y^2 + \log_2 z = 24
$$
Combine the logs:
$$
\log_2 (x^2 y^2 z) = 24
$$
Which means:
$$
x^2 y^2 z = 2^{24}
$$
Hmm, okay. So, I have the equation:
$$
x^2 y^2 z = 2^{24}
$$
We find the minimum value of (x + y + z). Since x, y, z are positive real numbers, I can use the AM-GM inequality to find the minimum.
But first, let me express x, y, z in terms of variables that can be handled by AM-GM. Maybe set variables a = x, b = y, c = z, but I need to adjust for the exponents.
Wait, the equation is x^2 y^2 z = 2^{24}. Let me write this as:
$$
(x y)^2 z = 2^{24}
$$
Let me set variables to make it symmetric. Let me let p = x y, q = z. Then:
$$
p^2 q = 2^{24}
$$
But I need to express x + y + z in terms of p and q. Hmm, maybe not the best approach.
Alternatively, let me consider the exponents. Let me take the logarithm base 2 of both sides:
$$
\log_2 (x^2 y^2 z) = 24
$$
Which is the same as:
$$
2 \log_2 x + 2 \log_2 y + \log_2 z = 24
$$
Let a = log_2 x, b = log_2 y, c = log_2 z. Then the equation becomes:
$$
2a + 2b + c = 24
$$
And we need to minimize x + y + z, which is 2^a + 2^b + 2^c.
So, the problem reduces to minimizing 2^a + 2^b + 2^c subject to 2a + 2b + c = 24, where a, b, c are real numbers.
This seems like a constrained optimization problem. I can use the method of Lagrange multipliers, but maybe there's a smarter way.
Alternatively, since all terms are positive, I can apply the AM-GM inequality.
But the exponents complicate things. Let u = 2a, v = 2b, w = c. Then the constraint becomes:
u + v + w = 24
And we need to minimize 2^{u/2} + 2^{v/2} + 2^{w}.
Hmm, not sure if that helps. Maybe another substitution.
Alternatively, let me consider that 2a + 2b + c = 24. Let me write this as:
2(a + b) + c = 24
Let me set d = a + b, so 2d + c = 24.
We need to minimize 2^a + 2^b + 2^c.
But I still have two variables a and b, which sum to d. To