Solving Weird Exponential Equations
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I am working on my math homework when I encountered a difficult problem. I simplified the equation and substituted smaller numbers to get this:
$n*2^n>10$
I have tried standard algebraic methods with logarithms, but I could not get them to work. Researching online, I came across the Lambert W function, but I know I don't need it to get the answer, as the math class I am taking is not that advanced. I strongly prefer not to use it.
If anyone can figure out the answer and explain, I would greatly appreciate it.
algebra-precalculus exponentiation
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show 4 more comments
$begingroup$
I am working on my math homework when I encountered a difficult problem. I simplified the equation and substituted smaller numbers to get this:
$n*2^n>10$
I have tried standard algebraic methods with logarithms, but I could not get them to work. Researching online, I came across the Lambert W function, but I know I don't need it to get the answer, as the math class I am taking is not that advanced. I strongly prefer not to use it.
If anyone can figure out the answer and explain, I would greatly appreciate it.
algebra-precalculus exponentiation
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If $n$ must be an integer, this equation holds for all $nge 3$.
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– vadim123
Jul 5 '14 at 23:58
1
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I think the only way to solve that equation is to use the Lambert W function or you could just try integers (if $n$ is an integer) to see what works. As @vadim123 states $ngeq 3$ since the actual solution is approximately: $n>2.1906$.
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– Jay
Jul 5 '14 at 23:58
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@vadim123, how did you figure that out?
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– alexwho314
Jul 6 '14 at 0:23
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@Jay, I would like to just try integers, but my teacher would probably prefer a better method than guess and check.
$endgroup$
– alexwho314
Jul 6 '14 at 0:25
2
$begingroup$
@alexwho314 I don't think there is any other method that's much better than guessing and checking. You could use various numerical methods but that's hardly better. I'm fairly sure there isn't another way of solving an equation of the form $x a^x=b$ with elementary functions. Unless you've made an error in getting that equation then I can't see an obvious way of doing this without guessing and checking (and without using the Lambert W function).
$endgroup$
– Jay
Jul 6 '14 at 0:31
|
show 4 more comments
$begingroup$
I am working on my math homework when I encountered a difficult problem. I simplified the equation and substituted smaller numbers to get this:
$n*2^n>10$
I have tried standard algebraic methods with logarithms, but I could not get them to work. Researching online, I came across the Lambert W function, but I know I don't need it to get the answer, as the math class I am taking is not that advanced. I strongly prefer not to use it.
If anyone can figure out the answer and explain, I would greatly appreciate it.
algebra-precalculus exponentiation
$endgroup$
I am working on my math homework when I encountered a difficult problem. I simplified the equation and substituted smaller numbers to get this:
$n*2^n>10$
I have tried standard algebraic methods with logarithms, but I could not get them to work. Researching online, I came across the Lambert W function, but I know I don't need it to get the answer, as the math class I am taking is not that advanced. I strongly prefer not to use it.
If anyone can figure out the answer and explain, I would greatly appreciate it.
algebra-precalculus exponentiation
algebra-precalculus exponentiation
asked Jul 5 '14 at 23:55
alexwho314alexwho314
132
132
$begingroup$
If $n$ must be an integer, this equation holds for all $nge 3$.
$endgroup$
– vadim123
Jul 5 '14 at 23:58
1
$begingroup$
I think the only way to solve that equation is to use the Lambert W function or you could just try integers (if $n$ is an integer) to see what works. As @vadim123 states $ngeq 3$ since the actual solution is approximately: $n>2.1906$.
$endgroup$
– Jay
Jul 5 '14 at 23:58
$begingroup$
@vadim123, how did you figure that out?
$endgroup$
– alexwho314
Jul 6 '14 at 0:23
$begingroup$
@Jay, I would like to just try integers, but my teacher would probably prefer a better method than guess and check.
$endgroup$
– alexwho314
Jul 6 '14 at 0:25
2
$begingroup$
@alexwho314 I don't think there is any other method that's much better than guessing and checking. You could use various numerical methods but that's hardly better. I'm fairly sure there isn't another way of solving an equation of the form $x a^x=b$ with elementary functions. Unless you've made an error in getting that equation then I can't see an obvious way of doing this without guessing and checking (and without using the Lambert W function).
$endgroup$
– Jay
Jul 6 '14 at 0:31
|
show 4 more comments
$begingroup$
If $n$ must be an integer, this equation holds for all $nge 3$.
$endgroup$
– vadim123
Jul 5 '14 at 23:58
1
$begingroup$
I think the only way to solve that equation is to use the Lambert W function or you could just try integers (if $n$ is an integer) to see what works. As @vadim123 states $ngeq 3$ since the actual solution is approximately: $n>2.1906$.
$endgroup$
– Jay
Jul 5 '14 at 23:58
$begingroup$
@vadim123, how did you figure that out?
$endgroup$
– alexwho314
Jul 6 '14 at 0:23
$begingroup$
@Jay, I would like to just try integers, but my teacher would probably prefer a better method than guess and check.
$endgroup$
– alexwho314
Jul 6 '14 at 0:25
2
$begingroup$
@alexwho314 I don't think there is any other method that's much better than guessing and checking. You could use various numerical methods but that's hardly better. I'm fairly sure there isn't another way of solving an equation of the form $x a^x=b$ with elementary functions. Unless you've made an error in getting that equation then I can't see an obvious way of doing this without guessing and checking (and without using the Lambert W function).
$endgroup$
– Jay
Jul 6 '14 at 0:31
$begingroup$
If $n$ must be an integer, this equation holds for all $nge 3$.
$endgroup$
– vadim123
Jul 5 '14 at 23:58
$begingroup$
If $n$ must be an integer, this equation holds for all $nge 3$.
$endgroup$
– vadim123
Jul 5 '14 at 23:58
1
1
$begingroup$
I think the only way to solve that equation is to use the Lambert W function or you could just try integers (if $n$ is an integer) to see what works. As @vadim123 states $ngeq 3$ since the actual solution is approximately: $n>2.1906$.
$endgroup$
– Jay
Jul 5 '14 at 23:58
$begingroup$
I think the only way to solve that equation is to use the Lambert W function or you could just try integers (if $n$ is an integer) to see what works. As @vadim123 states $ngeq 3$ since the actual solution is approximately: $n>2.1906$.
$endgroup$
– Jay
Jul 5 '14 at 23:58
$begingroup$
@vadim123, how did you figure that out?
$endgroup$
– alexwho314
Jul 6 '14 at 0:23
$begingroup$
@vadim123, how did you figure that out?
$endgroup$
– alexwho314
Jul 6 '14 at 0:23
$begingroup$
@Jay, I would like to just try integers, but my teacher would probably prefer a better method than guess and check.
$endgroup$
– alexwho314
Jul 6 '14 at 0:25
$begingroup$
@Jay, I would like to just try integers, but my teacher would probably prefer a better method than guess and check.
$endgroup$
– alexwho314
Jul 6 '14 at 0:25
2
2
$begingroup$
@alexwho314 I don't think there is any other method that's much better than guessing and checking. You could use various numerical methods but that's hardly better. I'm fairly sure there isn't another way of solving an equation of the form $x a^x=b$ with elementary functions. Unless you've made an error in getting that equation then I can't see an obvious way of doing this without guessing and checking (and without using the Lambert W function).
$endgroup$
– Jay
Jul 6 '14 at 0:31
$begingroup$
@alexwho314 I don't think there is any other method that's much better than guessing and checking. You could use various numerical methods but that's hardly better. I'm fairly sure there isn't another way of solving an equation of the form $x a^x=b$ with elementary functions. Unless you've made an error in getting that equation then I can't see an obvious way of doing this without guessing and checking (and without using the Lambert W function).
$endgroup$
– Jay
Jul 6 '14 at 0:31
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Do you remember how to graph exponential and reciprocal functions by hand? If so, rearrange to:
$$2^n > frac{10}{n}$$
The inequality is reversed if $n<0$, but it is obvious that no such solutions exist, so we ignore that case.
Then sketch the graphs by hand, note the intersection is around $n=2$, and test values $n=2$ and $n=3$ manually.
If you don't remember how to graph exponential and reciprocal functions, you can use numerical methods to approximate the point of intersection. Near zero, $2^n approx ln{2} cdot (1+n+frac{n^2}{2})$ (the first three terms of the Maclaurin sequence).
Using the well known approximation $ln{2} approx 0.7$, we can find the intersection by solving the polynomial:
$$
begin{equation}
begin{split}
10 &approx 0.7 left(n+n^2+frac{n^3}{2}right) newline
0 &approx 7n^3 + 14n^2 + 14n - 200
end{split}
end{equation}
$$
Use a standard scientific calculator to solve the above, to obtain $n approx 2.3$.
We then, once again, check $n=2$ and $n=3$ manually.
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$begingroup$
Thank you very much. As much as I would like to avoid a graphical method of solving, that must be what I have to do.
$endgroup$
– alexwho314
Jul 6 '14 at 15:55
add a comment |
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1 Answer
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$begingroup$
Do you remember how to graph exponential and reciprocal functions by hand? If so, rearrange to:
$$2^n > frac{10}{n}$$
The inequality is reversed if $n<0$, but it is obvious that no such solutions exist, so we ignore that case.
Then sketch the graphs by hand, note the intersection is around $n=2$, and test values $n=2$ and $n=3$ manually.
If you don't remember how to graph exponential and reciprocal functions, you can use numerical methods to approximate the point of intersection. Near zero, $2^n approx ln{2} cdot (1+n+frac{n^2}{2})$ (the first three terms of the Maclaurin sequence).
Using the well known approximation $ln{2} approx 0.7$, we can find the intersection by solving the polynomial:
$$
begin{equation}
begin{split}
10 &approx 0.7 left(n+n^2+frac{n^3}{2}right) newline
0 &approx 7n^3 + 14n^2 + 14n - 200
end{split}
end{equation}
$$
Use a standard scientific calculator to solve the above, to obtain $n approx 2.3$.
We then, once again, check $n=2$ and $n=3$ manually.
$endgroup$
$begingroup$
Thank you very much. As much as I would like to avoid a graphical method of solving, that must be what I have to do.
$endgroup$
– alexwho314
Jul 6 '14 at 15:55
add a comment |
$begingroup$
Do you remember how to graph exponential and reciprocal functions by hand? If so, rearrange to:
$$2^n > frac{10}{n}$$
The inequality is reversed if $n<0$, but it is obvious that no such solutions exist, so we ignore that case.
Then sketch the graphs by hand, note the intersection is around $n=2$, and test values $n=2$ and $n=3$ manually.
If you don't remember how to graph exponential and reciprocal functions, you can use numerical methods to approximate the point of intersection. Near zero, $2^n approx ln{2} cdot (1+n+frac{n^2}{2})$ (the first three terms of the Maclaurin sequence).
Using the well known approximation $ln{2} approx 0.7$, we can find the intersection by solving the polynomial:
$$
begin{equation}
begin{split}
10 &approx 0.7 left(n+n^2+frac{n^3}{2}right) newline
0 &approx 7n^3 + 14n^2 + 14n - 200
end{split}
end{equation}
$$
Use a standard scientific calculator to solve the above, to obtain $n approx 2.3$.
We then, once again, check $n=2$ and $n=3$ manually.
$endgroup$
$begingroup$
Thank you very much. As much as I would like to avoid a graphical method of solving, that must be what I have to do.
$endgroup$
– alexwho314
Jul 6 '14 at 15:55
add a comment |
$begingroup$
Do you remember how to graph exponential and reciprocal functions by hand? If so, rearrange to:
$$2^n > frac{10}{n}$$
The inequality is reversed if $n<0$, but it is obvious that no such solutions exist, so we ignore that case.
Then sketch the graphs by hand, note the intersection is around $n=2$, and test values $n=2$ and $n=3$ manually.
If you don't remember how to graph exponential and reciprocal functions, you can use numerical methods to approximate the point of intersection. Near zero, $2^n approx ln{2} cdot (1+n+frac{n^2}{2})$ (the first three terms of the Maclaurin sequence).
Using the well known approximation $ln{2} approx 0.7$, we can find the intersection by solving the polynomial:
$$
begin{equation}
begin{split}
10 &approx 0.7 left(n+n^2+frac{n^3}{2}right) newline
0 &approx 7n^3 + 14n^2 + 14n - 200
end{split}
end{equation}
$$
Use a standard scientific calculator to solve the above, to obtain $n approx 2.3$.
We then, once again, check $n=2$ and $n=3$ manually.
$endgroup$
Do you remember how to graph exponential and reciprocal functions by hand? If so, rearrange to:
$$2^n > frac{10}{n}$$
The inequality is reversed if $n<0$, but it is obvious that no such solutions exist, so we ignore that case.
Then sketch the graphs by hand, note the intersection is around $n=2$, and test values $n=2$ and $n=3$ manually.
If you don't remember how to graph exponential and reciprocal functions, you can use numerical methods to approximate the point of intersection. Near zero, $2^n approx ln{2} cdot (1+n+frac{n^2}{2})$ (the first three terms of the Maclaurin sequence).
Using the well known approximation $ln{2} approx 0.7$, we can find the intersection by solving the polynomial:
$$
begin{equation}
begin{split}
10 &approx 0.7 left(n+n^2+frac{n^3}{2}right) newline
0 &approx 7n^3 + 14n^2 + 14n - 200
end{split}
end{equation}
$$
Use a standard scientific calculator to solve the above, to obtain $n approx 2.3$.
We then, once again, check $n=2$ and $n=3$ manually.
edited Jul 6 '14 at 3:31
answered Jul 6 '14 at 2:50
Fengyang WangFengyang Wang
1,2741920
1,2741920
$begingroup$
Thank you very much. As much as I would like to avoid a graphical method of solving, that must be what I have to do.
$endgroup$
– alexwho314
Jul 6 '14 at 15:55
add a comment |
$begingroup$
Thank you very much. As much as I would like to avoid a graphical method of solving, that must be what I have to do.
$endgroup$
– alexwho314
Jul 6 '14 at 15:55
$begingroup$
Thank you very much. As much as I would like to avoid a graphical method of solving, that must be what I have to do.
$endgroup$
– alexwho314
Jul 6 '14 at 15:55
$begingroup$
Thank you very much. As much as I would like to avoid a graphical method of solving, that must be what I have to do.
$endgroup$
– alexwho314
Jul 6 '14 at 15:55
add a comment |
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$begingroup$
If $n$ must be an integer, this equation holds for all $nge 3$.
$endgroup$
– vadim123
Jul 5 '14 at 23:58
1
$begingroup$
I think the only way to solve that equation is to use the Lambert W function or you could just try integers (if $n$ is an integer) to see what works. As @vadim123 states $ngeq 3$ since the actual solution is approximately: $n>2.1906$.
$endgroup$
– Jay
Jul 5 '14 at 23:58
$begingroup$
@vadim123, how did you figure that out?
$endgroup$
– alexwho314
Jul 6 '14 at 0:23
$begingroup$
@Jay, I would like to just try integers, but my teacher would probably prefer a better method than guess and check.
$endgroup$
– alexwho314
Jul 6 '14 at 0:25
2
$begingroup$
@alexwho314 I don't think there is any other method that's much better than guessing and checking. You could use various numerical methods but that's hardly better. I'm fairly sure there isn't another way of solving an equation of the form $x a^x=b$ with elementary functions. Unless you've made an error in getting that equation then I can't see an obvious way of doing this without guessing and checking (and without using the Lambert W function).
$endgroup$
– Jay
Jul 6 '14 at 0:31