Mixing
Logs and indices questions: $4^{2x}-2^{x+1}=48$ and $6^{2x+1}-17*{6^x}+12=0$
$begingroup$
Hi can anyone solve these two questions using logs and indices
a.
$$4^{2x}-2^{x+1}=48$$
b.
$$6^{2x+1}-17*{6^x}+12=0$$
Thanks.
algebra-precalculus logarithms quadratics
edited Jan 23 at 7:54
Martin Sleziak
44.8k10119273
asked Sep 16 '14 at 12:29
Brass2010Brass2010
1255
$endgroup$
add a comment |
$begingroup$
Hi can anyone solve these two questions using logs and indices
a.
$$4^{2x}-2^{x+1}=48$$
b.
$$6^{2x+1}-17*{6^x}+12=0$$
Thanks.
algebra-precalculus logarithms quadratics
edited Jan 23 at 7:54
Martin Sleziak
44.8k10119273
asked Sep 16 '14 at 12:29
Brass2010Brass2010
1255
$endgroup$
$begingroup$
Is there a transcription error in the second (currently $6^{2x+1}-17(6x)+12=0$)? As it is you've got something that is not solvable, and not in a relatively simple form... Also, what have you tried?
$endgroup$
– Dan Uznanski
Sep 16 '14 at 12:32
add a comment |
$begingroup$
Hi can anyone solve these two questions using logs and indices
a.
$$4^{2x}-2^{x+1}=48$$
b.
$$6^{2x+1}-17*{6^x}+12=0$$
Thanks.
algebra-precalculus logarithms quadratics
edited Jan 23 at 7:54
Martin Sleziak
44.8k10119273
asked Sep 16 '14 at 12:29
Brass2010Brass2010
1255
$endgroup$
Hi can anyone solve these two questions using logs and indices
a.
$$4^{2x}-2^{x+1}=48$$
b.
$$6^{2x+1}-17*{6^x}+12=0$$
Thanks.
algebra-precalculus logarithms quadratics
algebra-precalculus logarithms quadratics
edited Jan 23 at 7:54
Martin Sleziak
44.8k10119273
asked Sep 16 '14 at 12:29
Brass2010Brass2010
1255
edited Jan 23 at 7:54
Martin Sleziak
44.8k10119273
asked Sep 16 '14 at 12:29
Brass2010Brass2010
1255
edited Jan 23 at 7:54
Martin Sleziak
44.8k10119273
edited Jan 23 at 7:54
Martin Sleziak
44.8k10119273
edited Jan 23 at 7:54
Martin Sleziak
44.8k10119273
44.8k10119273
asked Sep 16 '14 at 12:29
Brass2010Brass2010
1255
asked Sep 16 '14 at 12:29
Brass2010Brass2010
1255
asked Sep 16 '14 at 12:29
Brass2010Brass2010
1255
1255
$begingroup$
Is there a transcription error in the second (currently $6^{2x+1}-17(6x)+12=0$)? As it is you've got something that is not solvable, and not in a relatively simple form... Also, what have you tried?
$endgroup$
– Dan Uznanski
Sep 16 '14 at 12:32
add a comment |
$begingroup$
Is there a transcription error in the second (currently $6^{2x+1}-17(6x)+12=0$)? As it is you've got something that is not solvable, and not in a relatively simple form... Also, what have you tried?
$endgroup$
– Dan Uznanski
Sep 16 '14 at 12:32
$begingroup$
Is there a transcription error in the second (currently $6^{2x+1}-17(6x)+12=0$)? As it is you've got something that is not solvable, and not in a relatively simple form... Also, what have you tried?
$endgroup$
– Dan Uznanski
Sep 16 '14 at 12:32
$begingroup$
Is there a transcription error in the second (currently $6^{2x+1}-17(6x)+12=0$)? As it is you've got something that is not solvable, and not in a relatively simple form... Also, what have you tried?
$endgroup$
– Dan Uznanski
Sep 16 '14 at 12:32
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I believe the last question to be $6^{2x+1}-17(6^x)+12=0$
$$iff6(6^x)^2-17(6^x)+12=0$$
$$6^x=frac{17pmsqrt{17^2-4cdot6cdot12}}{2cdot6}=frac{17pm1}{12}=frac32,frac43$$
answered Sep 16 '14 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
I edited the question and that wasn't clear, but you should be correct. Thanks
$endgroup$
– cjferes
Sep 16 '14 at 12:39
$begingroup$
@cjferes, Thanks. I doubt the accuracy of the first question.
$endgroup$
– lab bhattacharjee
Sep 16 '14 at 12:46
$begingroup$
I guess, the first equation should be $4^x - 2^{x+1}=48,;$which leads to $(2^x)^2 -2cdot 2^x -48=0;$ with a solution $2^x=8;$ or $x=3.$
$endgroup$
– gammatester
Sep 16 '14 at 13:16
$begingroup$
the first question is indeed misleading, but i transcripted everything as the OP posted it.
$endgroup$
– cjferes
Sep 16 '14 at 13:38
add a comment |
$begingroup$
I am afraid but I do not think that there is any explicit solution for $$4^{2x}-2^{x+1}=48$$ To visualize it better, since the terms grow very fast, it is better to look at function $$f(x)=log(4^{2x})-log(2^{x+1}+48)=2xlog(4)-log(2^{x+1}+48)$$ which is basically a straight line.
To solve this equation, let us use Newton, which, starting from a reasonable guess $x_0$, will update it according to $$x_{n+1}=x_n-frac{f(x_n)}{f'(x_n)}$$ Here, we shall have $$f'(x)=2 log (4)-frac{2^{x+1} log (2)}{2^{x+1}+48}$$ Since the function seems to be very close to linearity, let me be lazy and start iterations at $x_0=0$. Then the successive iterates of Newton are $1.42522$, $1.43474$ which is the solution for six significant figures.
What is amazing is that, if you start the calculations at $x_0=1$, the result of the first iteration is $$x_1=frac{1}{51} left(25+frac{13 log (13)}{log (2)}right)approx 1.43345$$.
I also thought that there are some typo's in the equation but I give you this just for illustration purposes of a simple numerical solution of a quite difficult problem.
answered Sep 16 '14 at 17:02
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
Thanks for the answers, after a night of thoughts, I came up with the idea of using quadratic equations and logarithms to solve the answers
Let 2x=a
a^2-2a-48=0
a=8 or a=-6(rejected)
2^x=3
log(2)8=3
x=3
Let 6x=a
6a^2-17a+12=0
a=1.5 or4/3
log(6)1.5=0.226
log(6)1.5=0.1606
x=0.226 or 0.1606
Cheers!
answered Sep 16 '14 at 23:19
Brass2010Brass2010
1255
$endgroup$
$begingroup$
Since $4 = 2^2$, $4^{2x} = (2^2)^{2x} = 2^{4x} = (2^x)^4$. If we make the substitution $a = 2^x$, then $4^{2x} - 2^{x + 1} = (2^x)^{4} - 2 cdot 2^x = a^4 - 2a$.
$endgroup$
– N. F. Taussig
Jan 23 at 11:38
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I believe the last question to be $6^{2x+1}-17(6^x)+12=0$
$$iff6(6^x)^2-17(6^x)+12=0$$
$$6^x=frac{17pmsqrt{17^2-4cdot6cdot12}}{2cdot6}=frac{17pm1}{12}=frac32,frac43$$
answered Sep 16 '14 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
I edited the question and that wasn't clear, but you should be correct. Thanks
$endgroup$
– cjferes
Sep 16 '14 at 12:39
$begingroup$
@cjferes, Thanks. I doubt the accuracy of the first question.
$endgroup$
– lab bhattacharjee
Sep 16 '14 at 12:46
$begingroup$
I guess, the first equation should be $4^x - 2^{x+1}=48,;$which leads to $(2^x)^2 -2cdot 2^x -48=0;$ with a solution $2^x=8;$ or $x=3.$
$endgroup$
– gammatester
Sep 16 '14 at 13:16
$begingroup$
the first question is indeed misleading, but i transcripted everything as the OP posted it.
$endgroup$
– cjferes
Sep 16 '14 at 13:38
add a comment |
$begingroup$
I believe the last question to be $6^{2x+1}-17(6^x)+12=0$
$$iff6(6^x)^2-17(6^x)+12=0$$
$$6^x=frac{17pmsqrt{17^2-4cdot6cdot12}}{2cdot6}=frac{17pm1}{12}=frac32,frac43$$
answered Sep 16 '14 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
I edited the question and that wasn't clear, but you should be correct. Thanks
$endgroup$
– cjferes
Sep 16 '14 at 12:39
$begingroup$
@cjferes, Thanks. I doubt the accuracy of the first question.
$endgroup$
– lab bhattacharjee
Sep 16 '14 at 12:46
$begingroup$
I guess, the first equation should be $4^x - 2^{x+1}=48,;$which leads to $(2^x)^2 -2cdot 2^x -48=0;$ with a solution $2^x=8;$ or $x=3.$
$endgroup$
– gammatester
Sep 16 '14 at 13:16
$begingroup$
the first question is indeed misleading, but i transcripted everything as the OP posted it.
$endgroup$
– cjferes
Sep 16 '14 at 13:38
add a comment |
$begingroup$
I believe the last question to be $6^{2x+1}-17(6^x)+12=0$
$$iff6(6^x)^2-17(6^x)+12=0$$
$$6^x=frac{17pmsqrt{17^2-4cdot6cdot12}}{2cdot6}=frac{17pm1}{12}=frac32,frac43$$
answered Sep 16 '14 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
I believe the last question to be $6^{2x+1}-17(6^x)+12=0$
$$iff6(6^x)^2-17(6^x)+12=0$$
$$6^x=frac{17pmsqrt{17^2-4cdot6cdot12}}{2cdot6}=frac{17pm1}{12}=frac32,frac43$$
answered Sep 16 '14 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
answered Sep 16 '14 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
answered Sep 16 '14 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
answered Sep 16 '14 at 12:32
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
$begingroup$
I edited the question and that wasn't clear, but you should be correct. Thanks
$endgroup$
– cjferes
Sep 16 '14 at 12:39
$begingroup$
@cjferes, Thanks. I doubt the accuracy of the first question.
$endgroup$
– lab bhattacharjee
Sep 16 '14 at 12:46
$begingroup$
I guess, the first equation should be $4^x - 2^{x+1}=48,;$which leads to $(2^x)^2 -2cdot 2^x -48=0;$ with a solution $2^x=8;$ or $x=3.$
$endgroup$
– gammatester
Sep 16 '14 at 13:16
$begingroup$
the first question is indeed misleading, but i transcripted everything as the OP posted it.
$endgroup$
– cjferes
Sep 16 '14 at 13:38
add a comment |
$begingroup$
I edited the question and that wasn't clear, but you should be correct. Thanks
$endgroup$
– cjferes
Sep 16 '14 at 12:39
$begingroup$
@cjferes, Thanks. I doubt the accuracy of the first question.
$endgroup$
– lab bhattacharjee
Sep 16 '14 at 12:46
$begingroup$
I guess, the first equation should be $4^x - 2^{x+1}=48,;$which leads to $(2^x)^2 -2cdot 2^x -48=0;$ with a solution $2^x=8;$ or $x=3.$
$endgroup$
– gammatester
Sep 16 '14 at 13:16
$begingroup$
the first question is indeed misleading, but i transcripted everything as the OP posted it.
$endgroup$
– cjferes
Sep 16 '14 at 13:38
$begingroup$
I edited the question and that wasn't clear, but you should be correct. Thanks
$endgroup$
– cjferes
Sep 16 '14 at 12:39
$begingroup$
I edited the question and that wasn't clear, but you should be correct. Thanks
$endgroup$
– cjferes
Sep 16 '14 at 12:39
$begingroup$
@cjferes, Thanks. I doubt the accuracy of the first question.
$endgroup$
– lab bhattacharjee
Sep 16 '14 at 12:46
$begingroup$
@cjferes, Thanks. I doubt the accuracy of the first question.
$endgroup$
– lab bhattacharjee
Sep 16 '14 at 12:46
$begingroup$
I guess, the first equation should be $4^x - 2^{x+1}=48,;$which leads to $(2^x)^2 -2cdot 2^x -48=0;$ with a solution $2^x=8;$ or $x=3.$
$endgroup$
– gammatester
Sep 16 '14 at 13:16
$begingroup$
I guess, the first equation should be $4^x - 2^{x+1}=48,;$which leads to $(2^x)^2 -2cdot 2^x -48=0;$ with a solution $2^x=8;$ or $x=3.$
$endgroup$
– gammatester
Sep 16 '14 at 13:16
$begingroup$
the first question is indeed misleading, but i transcripted everything as the OP posted it.
$endgroup$
– cjferes
Sep 16 '14 at 13:38
$begingroup$
the first question is indeed misleading, but i transcripted everything as the OP posted it.
$endgroup$
– cjferes
Sep 16 '14 at 13:38
add a comment |
$begingroup$
I am afraid but I do not think that there is any explicit solution for $$4^{2x}-2^{x+1}=48$$ To visualize it better, since the terms grow very fast, it is better to look at function $$f(x)=log(4^{2x})-log(2^{x+1}+48)=2xlog(4)-log(2^{x+1}+48)$$ which is basically a straight line.
To solve this equation, let us use Newton, which, starting from a reasonable guess $x_0$, will update it according to $$x_{n+1}=x_n-frac{f(x_n)}{f'(x_n)}$$ Here, we shall have $$f'(x)=2 log (4)-frac{2^{x+1} log (2)}{2^{x+1}+48}$$ Since the function seems to be very close to linearity, let me be lazy and start iterations at $x_0=0$. Then the successive iterates of Newton are $1.42522$, $1.43474$ which is the solution for six significant figures.
What is amazing is that, if you start the calculations at $x_0=1$, the result of the first iteration is $$x_1=frac{1}{51} left(25+frac{13 log (13)}{log (2)}right)approx 1.43345$$.
I also thought that there are some typo's in the equation but I give you this just for illustration purposes of a simple numerical solution of a quite difficult problem.
answered Sep 16 '14 at 17:02
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
I am afraid but I do not think that there is any explicit solution for $$4^{2x}-2^{x+1}=48$$ To visualize it better, since the terms grow very fast, it is better to look at function $$f(x)=log(4^{2x})-log(2^{x+1}+48)=2xlog(4)-log(2^{x+1}+48)$$ which is basically a straight line.
To solve this equation, let us use Newton, which, starting from a reasonable guess $x_0$, will update it according to $$x_{n+1}=x_n-frac{f(x_n)}{f'(x_n)}$$ Here, we shall have $$f'(x)=2 log (4)-frac{2^{x+1} log (2)}{2^{x+1}+48}$$ Since the function seems to be very close to linearity, let me be lazy and start iterations at $x_0=0$. Then the successive iterates of Newton are $1.42522$, $1.43474$ which is the solution for six significant figures.
What is amazing is that, if you start the calculations at $x_0=1$, the result of the first iteration is $$x_1=frac{1}{51} left(25+frac{13 log (13)}{log (2)}right)approx 1.43345$$.
I also thought that there are some typo's in the equation but I give you this just for illustration purposes of a simple numerical solution of a quite difficult problem.
answered Sep 16 '14 at 17:02
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
I am afraid but I do not think that there is any explicit solution for $$4^{2x}-2^{x+1}=48$$ To visualize it better, since the terms grow very fast, it is better to look at function $$f(x)=log(4^{2x})-log(2^{x+1}+48)=2xlog(4)-log(2^{x+1}+48)$$ which is basically a straight line.
To solve this equation, let us use Newton, which, starting from a reasonable guess $x_0$, will update it according to $$x_{n+1}=x_n-frac{f(x_n)}{f'(x_n)}$$ Here, we shall have $$f'(x)=2 log (4)-frac{2^{x+1} log (2)}{2^{x+1}+48}$$ Since the function seems to be very close to linearity, let me be lazy and start iterations at $x_0=0$. Then the successive iterates of Newton are $1.42522$, $1.43474$ which is the solution for six significant figures.
What is amazing is that, if you start the calculations at $x_0=1$, the result of the first iteration is $$x_1=frac{1}{51} left(25+frac{13 log (13)}{log (2)}right)approx 1.43345$$.
I also thought that there are some typo's in the equation but I give you this just for illustration purposes of a simple numerical solution of a quite difficult problem.
answered Sep 16 '14 at 17:02
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
I am afraid but I do not think that there is any explicit solution for $$4^{2x}-2^{x+1}=48$$ To visualize it better, since the terms grow very fast, it is better to look at function $$f(x)=log(4^{2x})-log(2^{x+1}+48)=2xlog(4)-log(2^{x+1}+48)$$ which is basically a straight line.
To solve this equation, let us use Newton, which, starting from a reasonable guess $x_0$, will update it according to $$x_{n+1}=x_n-frac{f(x_n)}{f'(x_n)}$$ Here, we shall have $$f'(x)=2 log (4)-frac{2^{x+1} log (2)}{2^{x+1}+48}$$ Since the function seems to be very close to linearity, let me be lazy and start iterations at $x_0=0$. Then the successive iterates of Newton are $1.42522$, $1.43474$ which is the solution for six significant figures.
What is amazing is that, if you start the calculations at $x_0=1$, the result of the first iteration is $$x_1=frac{1}{51} left(25+frac{13 log (13)}{log (2)}right)approx 1.43345$$.
I also thought that there are some typo's in the equation but I give you this just for illustration purposes of a simple numerical solution of a quite difficult problem.
answered Sep 16 '14 at 17:02
Claude LeiboviciClaude Leibovici
124k1157135
answered Sep 16 '14 at 17:02
Claude LeiboviciClaude Leibovici
124k1157135
answered Sep 16 '14 at 17:02
Claude LeiboviciClaude Leibovici
124k1157135
answered Sep 16 '14 at 17:02
Claude LeiboviciClaude Leibovici
124k1157135
124k1157135
add a comment |
add a comment |
$begingroup$
Thanks for the answers, after a night of thoughts, I came up with the idea of using quadratic equations and logarithms to solve the answers
Let 2x=a
a^2-2a-48=0
a=8 or a=-6(rejected)
2^x=3
log(2)8=3
x=3
Let 6x=a
6a^2-17a+12=0
a=1.5 or4/3
log(6)1.5=0.226
log(6)1.5=0.1606
x=0.226 or 0.1606
Cheers!
answered Sep 16 '14 at 23:19
Brass2010Brass2010
1255
$endgroup$
$begingroup$
Since $4 = 2^2$, $4^{2x} = (2^2)^{2x} = 2^{4x} = (2^x)^4$. If we make the substitution $a = 2^x$, then $4^{2x} - 2^{x + 1} = (2^x)^{4} - 2 cdot 2^x = a^4 - 2a$.
$endgroup$
– N. F. Taussig
Jan 23 at 11:38
add a comment |
$begingroup$
Thanks for the answers, after a night of thoughts, I came up with the idea of using quadratic equations and logarithms to solve the answers
Let 2x=a
a^2-2a-48=0
a=8 or a=-6(rejected)
2^x=3
log(2)8=3
x=3
Let 6x=a
6a^2-17a+12=0
a=1.5 or4/3
log(6)1.5=0.226
log(6)1.5=0.1606
x=0.226 or 0.1606
Cheers!
answered Sep 16 '14 at 23:19
Brass2010Brass2010
1255
$endgroup$
$begingroup$
Since $4 = 2^2$, $4^{2x} = (2^2)^{2x} = 2^{4x} = (2^x)^4$. If we make the substitution $a = 2^x$, then $4^{2x} - 2^{x + 1} = (2^x)^{4} - 2 cdot 2^x = a^4 - 2a$.
$endgroup$
– N. F. Taussig
Jan 23 at 11:38
add a comment |
$begingroup$
Thanks for the answers, after a night of thoughts, I came up with the idea of using quadratic equations and logarithms to solve the answers
Let 2x=a
a^2-2a-48=0
a=8 or a=-6(rejected)
2^x=3
log(2)8=3
x=3
Let 6x=a
6a^2-17a+12=0
a=1.5 or4/3
log(6)1.5=0.226
log(6)1.5=0.1606
x=0.226 or 0.1606
Cheers!
answered Sep 16 '14 at 23:19
Brass2010Brass2010
1255
$endgroup$
Thanks for the answers, after a night of thoughts, I came up with the idea of using quadratic equations and logarithms to solve the answers
Let 2x=a
a^2-2a-48=0
a=8 or a=-6(rejected)
2^x=3
log(2)8=3
x=3
Let 6x=a
6a^2-17a+12=0
a=1.5 or4/3
log(6)1.5=0.226
log(6)1.5=0.1606
x=0.226 or 0.1606
Cheers!
answered Sep 16 '14 at 23:19
Brass2010Brass2010
1255
answered Sep 16 '14 at 23:19
Brass2010Brass2010
1255
answered Sep 16 '14 at 23:19
Brass2010Brass2010
1255
answered Sep 16 '14 at 23:19
Brass2010Brass2010
1255
1255
$begingroup$
Since $4 = 2^2$, $4^{2x} = (2^2)^{2x} = 2^{4x} = (2^x)^4$. If we make the substitution $a = 2^x$, then $4^{2x} - 2^{x + 1} = (2^x)^{4} - 2 cdot 2^x = a^4 - 2a$.
$endgroup$
– N. F. Taussig
Jan 23 at 11:38
add a comment |
$begingroup$
Since $4 = 2^2$, $4^{2x} = (2^2)^{2x} = 2^{4x} = (2^x)^4$. If we make the substitution $a = 2^x$, then $4^{2x} - 2^{x + 1} = (2^x)^{4} - 2 cdot 2^x = a^4 - 2a$.
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– N. F. Taussig
Jan 23 at 11:38
$begingroup$
Since $4 = 2^2$, $4^{2x} = (2^2)^{2x} = 2^{4x} = (2^x)^4$. If we make the substitution $a = 2^x$, then $4^{2x} - 2^{x + 1} = (2^x)^{4} - 2 cdot 2^x = a^4 - 2a$.
$endgroup$
– N. F. Taussig
Jan 23 at 11:38
$begingroup$
Since $4 = 2^2$, $4^{2x} = (2^2)^{2x} = 2^{4x} = (2^x)^4$. If we make the substitution $a = 2^x$, then $4^{2x} - 2^{x + 1} = (2^x)^{4} - 2 cdot 2^x = a^4 - 2a$.
$endgroup$
– N. F. Taussig
Jan 23 at 11:38
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$begingroup$
Is there a transcription error in the second (currently $6^{2x+1}-17(6x)+12=0$)? As it is you've got something that is not solvable, and not in a relatively simple form... Also, what have you tried?
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– Dan Uznanski
Sep 16 '14 at 12:32