If 1212x×424x+12×5^{2y} = 8^{4z}×2012x×243^{3x-6} , where x,y and z are natural numbers, then x+y+z equals
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Official Correct Answer: 112. The equation given is:
\(12^{12x} \times 4^{24x+12} \times 5^{2y} = 8^{4z} \times 20^{12x} \times 243^{3x-6}\)
And we need to find \(x + y + z\) where \(x\), \(y\), and \(z\) are natural numbers. The correct answer is supposed to be 112, so I need to make sure I get there.
First, when dealing with equations involving exponents, it's helpful to express all the bases as powers of prime numbers. That way, I can equate the exponents of the same primes on both sides of the equation. The prime numbers involved here seem to be 2, 3, and 5. Let me rewrite each term accordingly.
Starting with the left-hand side (LHS):
1. \(12^{12x}\): 12 can be written as \(2^2 \times 3^1\). So, \(12^{12x} = (2^2 \times 3)^{12x} = 2^{24x} \times 3^{12x}\).
2. \(4^{24x+12}\): 4 is \(2^2\), so this becomes \((2^2)^{24x+12} = 2^{48x + 24}\).
3. \(5^{2y}\) remains as it is because 5 is already a prime.
So, putting it all together, the LHS becomes:
\(2^{24x} \times 3^{12x} \times 2^{48x + 24} \times 5^{2y}\)
I can combine the exponents of the same base on the LHS:
- For base 2: \(24x + 48x + 24 = 72x + 24\)
- For base 3: \(12x\)
- For base 5: \(2y\)
So, LHS simplifies to:
\(2^{72x + 24} \times 3^{12x} \times 5^{2y}\)
Now, moving on to the right-hand side (RHS):
1. \(8^{4z}\): 8 is \(2^3\), so this becomes \((2^3)^{4z} = 2^{12z}\).
2. \(20^{12x}\): 20 is \(2^2 \times 5^1\), so this becomes \((2^2 \times 5)^{12x} = 2^{24x} \times 5^{12x}\).
3. \(243^{3x-6}\): 243 is \(3^5\), so this becomes \((3^5)^{3x - 6} = 3^{15x - 30}\).
Putting it all together, the RHS becomes:
\(2^{12z} \times 2^{24x} \times 5^{12x} \times 3^{15x - 30}\)
Again, combining the exponents of the same base on the RHS:
- For base 2: \(12z + 24x\)
- For base 3: \(15x - 30\)
- For base 5: \(12x\)
So, RHS simplifies to:
\(2^{24x + 12z} \times 3^{15x - 30} \times 5^{12x}\)
Now, since the LHS and RHS must be equal, their exponents for each prime base must be equal. So, I can set up equations for each prime:
1. For base 2:
\(72x + 24 = 24x + 12z\)
2. For base 3:
\(12x = 15x - 30\)
3. For base 5:
\(2y = 12x\)
Let me solve these equations one by one.
Starting with base 3:
\(12x = 15x - 30\)
Subtract \(12x\) from both sides:
\(0 = 3x - 30\)
Add 30 to both sides:
\(30 = 3x\)
Divide both sides by 3:
\(x = 10\)
Okay, so \(x = 10\). That was straightforward.
Next, let's use this value of \(x\) in the equation for base 5:
\(2y = 12x\)
Substitute \(x = 10\):
\(2y = 12 \times 10 = 120\)
Divide both sides by 2:
\(y = 60\)
Great, so \(y = 60\).
Now, moving on to the equation for base 2:
\(72x + 24 = 24x + 12z\)
Substitute \(x = 10\):
\(72 \times 10 + 24 = 24 \times 10 + 12z\)
Calculate each term:
\(720 + 24 = 240 + 12z\)
So,
\(744 = 240 + 12z\)
Subtract 240 from both sides:
\(744 - 240 = 12z\)
\(504 = 12z\)
Divide both sides by 12:
\(z = 42\)
Alright, so \(z = 42\).
Now, we have \(x = 10\), \(y = 60\), and \(z = 42\). Adding them together:
\(x + y + z = 10 + 60 + 42 = 112\)
So, the sum is 112. That matches the given correct answer. I think I did everything correctly.
- Expressed all terms as prime factors correctly.
- Combined exponents on both sides accurately.
- Solved each equation step by step without errors.
- Substituted values correctly and performed arithmetic accurately.
Everything seems in order. I'm confident that the answer is indeed 112.
**Final Answer**
The value of \(x + y + z\) is \