Mixing
Help rearranging algebraic expression?!
$begingroup$
In the following relationship x and $delta_S$ are unknowns:
$$delta_f = delta_S-(delta_S-delta_o)expleft[-(1+mx)frac{It}{V}right]$$
I need to solve for x and have the following relationship for $delta_S$:
$$delta_S =frac{delta_{in}+mxdelta^*}{1+mx}$$
Therefore: $$delta_f = bigg(frac{(delta_{in}+mxdelta^*)}{(1+mx)}bigg)-bigg(frac{(delta_{in}+mxdelta^*)}{(1+mx)}-delta_ibigg)expbigg[-(1+mx)frac{It}{V}bigg]$$
Honestly I have no idea where to begin isolating x once I start taking the natural log...
logarithms exponential-function factoring
edited Jan 19 at 1:58
Larry
2,41331129
asked Jan 19 at 0:51
spencerchadspencerchad
82
$endgroup$
add a comment |
$begingroup$
In the following relationship x and $delta_S$ are unknowns:
$$delta_f = delta_S-(delta_S-delta_o)expleft[-(1+mx)frac{It}{V}right]$$
I need to solve for x and have the following relationship for $delta_S$:
$$delta_S =frac{delta_{in}+mxdelta^*}{1+mx}$$
Therefore: $$delta_f = bigg(frac{(delta_{in}+mxdelta^*)}{(1+mx)}bigg)-bigg(frac{(delta_{in}+mxdelta^*)}{(1+mx)}-delta_ibigg)expbigg[-(1+mx)frac{It}{V}bigg]$$
Honestly I have no idea where to begin isolating x once I start taking the natural log...
logarithms exponential-function factoring
edited Jan 19 at 1:58
Larry
2,41331129
asked Jan 19 at 0:51
spencerchadspencerchad
82
$endgroup$
add a comment |
$begingroup$
In the following relationship x and $delta_S$ are unknowns:
$$delta_f = delta_S-(delta_S-delta_o)expleft[-(1+mx)frac{It}{V}right]$$
I need to solve for x and have the following relationship for $delta_S$:
$$delta_S =frac{delta_{in}+mxdelta^*}{1+mx}$$
Therefore: $$delta_f = bigg(frac{(delta_{in}+mxdelta^*)}{(1+mx)}bigg)-bigg(frac{(delta_{in}+mxdelta^*)}{(1+mx)}-delta_ibigg)expbigg[-(1+mx)frac{It}{V}bigg]$$
Honestly I have no idea where to begin isolating x once I start taking the natural log...
logarithms exponential-function factoring
edited Jan 19 at 1:58
Larry
2,41331129
asked Jan 19 at 0:51
spencerchadspencerchad
82
$endgroup$
In the following relationship x and $delta_S$ are unknowns:
$$delta_f = delta_S-(delta_S-delta_o)expleft[-(1+mx)frac{It}{V}right]$$
I need to solve for x and have the following relationship for $delta_S$:
$$delta_S =frac{delta_{in}+mxdelta^*}{1+mx}$$
Therefore: $$delta_f = bigg(frac{(delta_{in}+mxdelta^*)}{(1+mx)}bigg)-bigg(frac{(delta_{in}+mxdelta^*)}{(1+mx)}-delta_ibigg)expbigg[-(1+mx)frac{It}{V}bigg]$$
Honestly I have no idea where to begin isolating x once I start taking the natural log...
logarithms exponential-function factoring
logarithms exponential-function factoring
edited Jan 19 at 1:58
Larry
2,41331129
asked Jan 19 at 0:51
spencerchadspencerchad
82
edited Jan 19 at 1:58
Larry
2,41331129
asked Jan 19 at 0:51
spencerchadspencerchad
82
edited Jan 19 at 1:58
Larry
2,41331129
edited Jan 19 at 1:58
Larry
2,41331129
edited Jan 19 at 1:58
Larry
2,41331129
2,41331129
asked Jan 19 at 0:51
spencerchadspencerchad
82
asked Jan 19 at 0:51
spencerchadspencerchad
82
asked Jan 19 at 0:51
spencerchadspencerchad
82
82
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
You cannot isolate $x$ and you will need some numerical method.
If I had to do it, I would define $y=(1+mx)$ to make
$$delta_f = delta_S-(delta_S-delta_o)expleft[-yfrac{It}{V}right]$$
$$delta_S =frac{delta_{in}+mxdelta^*}{1+mx}=frac{delta_{in}-delta^*+ydelta^*}{y}=delta^*+frac{delta_{in}-delta^*}y$$ Use Newton method.
In fact, I suppose that there could be an "explicit" solution in terms of the generalized Lambert function (have a look at equation $(4)$ in the paper).
answered Jan 19 at 4:18
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
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$begingroup$
You cannot isolate $x$ and you will need some numerical method.
If I had to do it, I would define $y=(1+mx)$ to make
$$delta_f = delta_S-(delta_S-delta_o)expleft[-yfrac{It}{V}right]$$
$$delta_S =frac{delta_{in}+mxdelta^*}{1+mx}=frac{delta_{in}-delta^*+ydelta^*}{y}=delta^*+frac{delta_{in}-delta^*}y$$ Use Newton method.
In fact, I suppose that there could be an "explicit" solution in terms of the generalized Lambert function (have a look at equation $(4)$ in the paper).
answered Jan 19 at 4:18
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
You cannot isolate $x$ and you will need some numerical method.
If I had to do it, I would define $y=(1+mx)$ to make
$$delta_f = delta_S-(delta_S-delta_o)expleft[-yfrac{It}{V}right]$$
$$delta_S =frac{delta_{in}+mxdelta^*}{1+mx}=frac{delta_{in}-delta^*+ydelta^*}{y}=delta^*+frac{delta_{in}-delta^*}y$$ Use Newton method.
In fact, I suppose that there could be an "explicit" solution in terms of the generalized Lambert function (have a look at equation $(4)$ in the paper).
answered Jan 19 at 4:18
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
You cannot isolate $x$ and you will need some numerical method.
If I had to do it, I would define $y=(1+mx)$ to make
$$delta_f = delta_S-(delta_S-delta_o)expleft[-yfrac{It}{V}right]$$
$$delta_S =frac{delta_{in}+mxdelta^*}{1+mx}=frac{delta_{in}-delta^*+ydelta^*}{y}=delta^*+frac{delta_{in}-delta^*}y$$ Use Newton method.
In fact, I suppose that there could be an "explicit" solution in terms of the generalized Lambert function (have a look at equation $(4)$ in the paper).
answered Jan 19 at 4:18
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
You cannot isolate $x$ and you will need some numerical method.
If I had to do it, I would define $y=(1+mx)$ to make
$$delta_f = delta_S-(delta_S-delta_o)expleft[-yfrac{It}{V}right]$$
$$delta_S =frac{delta_{in}+mxdelta^*}{1+mx}=frac{delta_{in}-delta^*+ydelta^*}{y}=delta^*+frac{delta_{in}-delta^*}y$$ Use Newton method.
In fact, I suppose that there could be an "explicit" solution in terms of the generalized Lambert function (have a look at equation $(4)$ in the paper).
answered Jan 19 at 4:18
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 4:18
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 4:18
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 4:18
Claude LeiboviciClaude Leibovici
123k1157134
123k1157134
add a comment |
add a comment |
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