[color=#ff0000] where do you see power of 3 ? 
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Difficulty:
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Question Stats:
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18%
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If x and k are integers and \((12^x)(4^{2x+ 1})= (2^k)(3^2)\), what is the value of k ?
(A) 5
(B) 7
(C) 10
(D) 12
(E) 14
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(A) 5
(B) 7
(C) 10
(D) 12
(E) 14
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SOLUTION
If x and k are integers and \((12^x)(4^{2x+ 1})= (2^k)(3^2)\), what is the value of k ?
(A) 5
(B) 7
(C) 10
(D) 12
(E) 14
\((12^x)(4^{2x+ 1})= (2^k)(3^2)\);
\((2^{2x}*3^x)(2^{2(2x+ 1)})= (2^k)(3^2)\);
\((2^{2x+2(2x+ 1)}*3^x)= (2^k)(3^2)\);
Equate the powers of 3 --> \(x=2\);
Equate the powers of 2 --> \(2x+2(2x+ 1)=k\) --> \(6x+2=k\) --> \(k=14\).
Answer: E.
_________________
If x and k are integers and \((12^x)(4^{2x+ 1})= (2^k)(3^2)\), what is the value of k ?
(A) 5
(B) 7
(C) 10
(D) 12
(E) 14
\((12^x)(4^{2x+ 1})= (2^k)(3^2)\);
\((2^{2x}*3^x)(2^{2(2x+ 1)})= (2^k)(3^2)\);
\((2^{2x+2(2x+ 1)}*3^x)= (2^k)(3^2)\);
Equate the powers of 3 --> \(x=2\);
Equate the powers of 2 --> \(2x+2(2x+ 1)=k\) --> \(6x+2=k\) --> \(k=14\).
Answer: E.
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Answer = (E) 14
Re-write as
\(3^x . 4^x . 4^{(2x+1)} = 2^k. 3^2\)
LHS & RHS has only 1 base of 3, so x = 2
\(4^2 . 4^5 = 2^k\)
\(4^7 = 2^k\)
K = 14
Re-write as
\(3^x . 4^x . 4^{(2x+1)} = 2^k. 3^2\)
LHS & RHS has only 1 base of 3, so x = 2
\(4^2 . 4^5 = 2^k\)
\(4^7 = 2^k\)
K = 14
General Discussion
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Posts: 520
If x and k are integers and \((12^x)(4^{2x+ 1})= (2^k)(3^2)\), what is the value of k ?
(A) 5
(B) 7
(C) 10
(D) 12
(E) 14
Sol: The given expression can be reduced to (3^x*4^x)4^(2x+1)= 2^k*3^2
So we see that 3^x=3^2 or x=2
Also 4^x*4^(2x+1) can be written as 4^3x+1 =2^k or 2^2(3x+1)= 2^k or k=2*(3x+1) or k=14
Ans is E
600 level is okay.
(A) 5
(B) 7
(C) 10
(D) 12
(E) 14
Sol: The given expression can be reduced to (3^x*4^x)4^(2x+1)= 2^k*3^2
So we see that 3^x=3^2 or x=2
Also 4^x*4^(2x+1) can be written as 4^3x+1 =2^k or 2^2(3x+1)= 2^k or k=2*(3x+1) or k=14
Ans is E
600 level is okay.
Its E..
12=3*2^2
4^2x+1=2^2(2x+1)
well will get x=2..
and 2x+4x+2=k
6x+2=k
x=2
k=14
12=3*2^2
4^2x+1=2^2(2x+1)
well will get x=2..
and 2x+4x+2=k
6x+2=k
x=2
k=14
JeffTargetTestPrep
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Expert reply
Posts: 2,973
Bunuel
Let’s simplify the given equation:
(3 * 2^2)^x * (2^2)^(2x + 1) = 2^k * 3^2
3^x * 2^(2x) * 2^(4x + 2) = 2^k * 3^2
3^x * 2^(6x + 2) = 3^2 * 2^k
Equating the exponents with the like bases, we see that:
x = 2 and 6x + 2 = k
Thus, k = 6(2) + 2 = 14.
Answer: E
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Bunuel
Hi Bunuel
\((2^{2x+2(2x+ 1)}*3^x)= (2^k)(3^2)\); after this
\((2^{6x+2}*3^x)= (2^k)(3^2)\); <--- i get this
how do you Equate the powers of 3 --> \(x=2\); :? where do you see power of 3 ? :? how you get 2 :?
how do you Equate the powers of 2 --> \(2x+2(2x+ 1)=k\) --> \(6x+2=k\) --> \(k=14\). and get 14 :?
i know this rule -->
if \(a^x = a^y\) then \(x = y\)
in other words if the bases match then exponents are equal.
please explain
\(thank you ^ {1000}\)
dave13
2^(2x+2(2x+ 1))*3^x= (2^k)*(3^2);
Equate the powers of 3 --> \(x=2\);
Next, equate the powers of 2: \(2x+2(2x+ 1)=k\) --> \(6x+2=k\). SINCE x = 2 (form above), then \(k=14\).
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Bunuel
Given: \((12^x)(4^{2x+ 1})= (2^k)(3^2)\)
Rewrite \(12\) and \(4\) as follows: \([(2^2)(3^1)]^x[(2^2)^{2x+ 1}]= (2^k)(3^2)\)
Apply exponent laws to get: \((2^{2x})(3^x)(2^{4x+ 2})= (2^k)(3^2)\)
On the left side, combine the powers of \(2\) to get: \((2^{6x+2})(3^x)= (2^k)(3^2)\)
At this point we can see that \(6x+2 = k\) AND \(x = 2\)
Take: \(6x+2 = k\)
Replace \(x\) with \(2\) to get: \(6(2)+2 = k\)
Evaluate: \(14 = k\)
Answer: E
Cheers,
Brent
Posts: 76
\(12^x4^{(2x+1)}= (2^k)(3^2)\)
\((3*2^2)^x(2^2)^{(2x+1)}= (2^k)(3^2)\)
\(3^x2^{x+2(2x+1)}= (2^k)(3^2)\)
\(x=2\)
\(k=2x+2(2x+1)\)
\(k=2*2+2(2*2+1)\)
\(k=14\)
Answer E
\((3*2^2)^x(2^2)^{(2x+1)}= (2^k)(3^2)\)
\(3^x2^{x+2(2x+1)}= (2^k)(3^2)\)
\(x=2\)
\(k=2x+2(2x+1)\)
\(k=2*2+2(2*2+1)\)
\(k=14\)
Answer E
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Asked: If x and k are integers and \((12^x)(4^{2x+ 1})= (2^k)(3^2)\), what is the value of k ?
\(3^x*2^{2x}*2^{4x+2} = 2^k*3^2\)
\(2^{6x+2}*3^x = 2^k*3^2\)
Equating the powers of 3 & 2 respectively, we get
x = 2
k = 6x+2 = 14
IMO E
\(3^x*2^{2x}*2^{4x+2} = 2^k*3^2\)
\(2^{6x+2}*3^x = 2^k*3^2\)
Equating the powers of 3 & 2 respectively, we get
x = 2
k = 6x+2 = 14
IMO E
Posts: 5
\((12^{x})(4^{2x+1}) = (2^{k})(3^{2})\), find value of k
Break down 12 and 4 into smaller parts
\((3*4^{x})(2*2)^{2x+1} = (2^{k})(3^{2})\)
\((3*2*2)^{x}(2*2)^{2x+1} = (2^{k})(3^{2})\)
Apply exponents
\((3^{x}*2^{x}*2^{x})*(2^{2x+1}*2^{2x+1}) = (2^{k})(3^{2})\)
Add exponents with same base together
\(3^{x}*2^{6x+2} = (2^{k})(3^{2})\)
Solve for base 3 to get x
\(3^{x}=3^{2}\)
\(x=2\)
Apply x to k formula
\(2^{k}=2^{6x+2}\)
\(k=6x+2\)
\(k=6(2)+2\)
\(k=14\)
Break down 12 and 4 into smaller parts
\((3*4^{x})(2*2)^{2x+1} = (2^{k})(3^{2})\)
\((3*2*2)^{x}(2*2)^{2x+1} = (2^{k})(3^{2})\)
Apply exponents
\((3^{x}*2^{x}*2^{x})*(2^{2x+1}*2^{2x+1}) = (2^{k})(3^{2})\)
Add exponents with same base together
\(3^{x}*2^{6x+2} = (2^{k})(3^{2})\)
Solve for base 3 to get x
\(3^{x}=3^{2}\)
\(x=2\)
Apply x to k formula
\(2^{k}=2^{6x+2}\)
\(k=6x+2\)
\(k=6(2)+2\)
\(k=14\)
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