Mixing
Solving Simultaneous Equations with logarithms (require all steps)
$begingroup$
I would really appreciate a full solution with all working out to the following simultaneous equations, as I can't seem to arrive at the same answer as a text book on part of a question.
The equations are:
$$ V_1 = A ln r_1 + B$$
$$ V_2 = A ln r_2 + B$$
The answer in the book is:
$$ A = frac{V_2 - V_1}{ln r_2 - ln r_1} = frac{V_2 - V_1}{ln (r_2/r_1)} $$
$$ B = frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} $$
Inserting these into the general equation $$ V(r) = A ln r + B$$ gives:
$$ V(r) = frac{V_2 - V_1}{ln (r_2/r_1)} ln r + frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} $$
$$ V(r) = frac{V_2 ln(r/r_1) - V_1 ln(r/r_2)}{ln (r_2/r_1)} $$
Thank you
logarithms systems-of-equations
edited Jan 28 at 12:44
Harry Peter
5,48911439
asked Jan 23 at 20:08
frejafreja
154
$endgroup$
add a comment |
$begingroup$
I would really appreciate a full solution with all working out to the following simultaneous equations, as I can't seem to arrive at the same answer as a text book on part of a question.
The equations are:
$$ V_1 = A ln r_1 + B$$
$$ V_2 = A ln r_2 + B$$
The answer in the book is:
$$ A = frac{V_2 - V_1}{ln r_2 - ln r_1} = frac{V_2 - V_1}{ln (r_2/r_1)} $$
$$ B = frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} $$
Inserting these into the general equation $$ V(r) = A ln r + B$$ gives:
$$ V(r) = frac{V_2 - V_1}{ln (r_2/r_1)} ln r + frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} $$
$$ V(r) = frac{V_2 ln(r/r_1) - V_1 ln(r/r_2)}{ln (r_2/r_1)} $$
Thank you
logarithms systems-of-equations
edited Jan 28 at 12:44
Harry Peter
5,48911439
asked Jan 23 at 20:08
frejafreja
154
$endgroup$
$begingroup$
What is wrong with the answer in the book? It is a simple solution to a simple problem. Your approach seems to be unnecessarily complicated.
$endgroup$
– herb steinberg
Jan 23 at 20:43
$begingroup$
What do you mean? The last line is the book answer, I don't understand how they got there.
$endgroup$
– freja
Jan 23 at 22:23
add a comment |
$begingroup$
I would really appreciate a full solution with all working out to the following simultaneous equations, as I can't seem to arrive at the same answer as a text book on part of a question.
The equations are:
$$ V_1 = A ln r_1 + B$$
$$ V_2 = A ln r_2 + B$$
The answer in the book is:
$$ A = frac{V_2 - V_1}{ln r_2 - ln r_1} = frac{V_2 - V_1}{ln (r_2/r_1)} $$
$$ B = frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} $$
Inserting these into the general equation $$ V(r) = A ln r + B$$ gives:
$$ V(r) = frac{V_2 - V_1}{ln (r_2/r_1)} ln r + frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} $$
$$ V(r) = frac{V_2 ln(r/r_1) - V_1 ln(r/r_2)}{ln (r_2/r_1)} $$
Thank you
logarithms systems-of-equations
edited Jan 28 at 12:44
Harry Peter
5,48911439
asked Jan 23 at 20:08
frejafreja
154
$endgroup$
I would really appreciate a full solution with all working out to the following simultaneous equations, as I can't seem to arrive at the same answer as a text book on part of a question.
The equations are:
$$ V_1 = A ln r_1 + B$$
$$ V_2 = A ln r_2 + B$$
The answer in the book is:
$$ A = frac{V_2 - V_1}{ln r_2 - ln r_1} = frac{V_2 - V_1}{ln (r_2/r_1)} $$
$$ B = frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} $$
Inserting these into the general equation $$ V(r) = A ln r + B$$ gives:
$$ V(r) = frac{V_2 - V_1}{ln (r_2/r_1)} ln r + frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} $$
$$ V(r) = frac{V_2 ln(r/r_1) - V_1 ln(r/r_2)}{ln (r_2/r_1)} $$
Thank you
logarithms systems-of-equations
logarithms systems-of-equations
edited Jan 28 at 12:44
Harry Peter
5,48911439
asked Jan 23 at 20:08
frejafreja
154
edited Jan 28 at 12:44
Harry Peter
5,48911439
asked Jan 23 at 20:08
frejafreja
154
edited Jan 28 at 12:44
Harry Peter
5,48911439
edited Jan 28 at 12:44
Harry Peter
5,48911439
edited Jan 28 at 12:44
Harry Peter
5,48911439
5,48911439
asked Jan 23 at 20:08
frejafreja
154
asked Jan 23 at 20:08
frejafreja
154
asked Jan 23 at 20:08
frejafreja
154
154
$begingroup$
What is wrong with the answer in the book? It is a simple solution to a simple problem. Your approach seems to be unnecessarily complicated.
$endgroup$
– herb steinberg
Jan 23 at 20:43
$begingroup$
What do you mean? The last line is the book answer, I don't understand how they got there.
$endgroup$
– freja
Jan 23 at 22:23
add a comment |
$begingroup$
What is wrong with the answer in the book? It is a simple solution to a simple problem. Your approach seems to be unnecessarily complicated.
$endgroup$
– herb steinberg
Jan 23 at 20:43
$begingroup$
What do you mean? The last line is the book answer, I don't understand how they got there.
$endgroup$
– freja
Jan 23 at 22:23
$begingroup$
What is wrong with the answer in the book? It is a simple solution to a simple problem. Your approach seems to be unnecessarily complicated.
$endgroup$
– herb steinberg
Jan 23 at 20:43
$begingroup$
What is wrong with the answer in the book? It is a simple solution to a simple problem. Your approach seems to be unnecessarily complicated.
$endgroup$
– herb steinberg
Jan 23 at 20:43
$begingroup$
What do you mean? The last line is the book answer, I don't understand how they got there.
$endgroup$
– freja
Jan 23 at 22:23
$begingroup$
What do you mean? The last line is the book answer, I don't understand how they got there.
$endgroup$
– freja
Jan 23 at 22:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your equations are
$$V_1=Aln r_1+Btag1$$
$$V_2=Aln r_2+Btag2$$
Subtracting $(1)$ from $(2)$ we obtain
$$V_2-V_1=A(ln r_2-ln r_1)$$
$$implies A=frac{V_2 - V_1}{ln r_2 - ln r_1} = frac{V_2 - V_1}{ln (r_2/r_1)}$$
Pluging this result into $(1)$ we get
$$begin{align}
V_1&=frac{V_2 - V_1}{ln (r_2/r_1)}ln r_1 + B \
implies B&=V_1-frac{V_2 - V_1}{ln (r_2/r_1)}ln r_1 \
&= frac{V_1 ln(r_2/r_1)-(V_2-V_1)ln r_1}{ln (r_2/r_1)} \
&=frac{V_1ln r_2-V_1ln r_1-V_2ln r_1+V_1ln r_1}{ln (r_2/r_1)} \
&=frac{V_1ln r_2 - V_2ln r_1}{ln (r_2/r_1)} \
end{align}$$
Inserting these into the equation $V(r)=Aln r+B$ gives
$$begin{align}
V(r) &= frac{V_2 - V_1}{ln (r_2/r_1)} ln r + frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} \
&=frac{V_2ln r - V_1ln r+V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} \
&=frac{V_2(ln r - ln r_1) - V_1(ln r - ln r_2)}{ln (r_2/r_1)} \
&=frac{V_2ln (r/r_1) - V_1ln (r/r_2)}{ln (r_2/r_1)}
end{align}$$
as desired.
answered Jan 23 at 20:54
greeliousgreelious
472112
$endgroup$
$begingroup$
That's great, thank you. Much appreciated, and makes sense.
$endgroup$
– freja
Jan 23 at 22:27
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your equations are
$$V_1=Aln r_1+Btag1$$
$$V_2=Aln r_2+Btag2$$
Subtracting $(1)$ from $(2)$ we obtain
$$V_2-V_1=A(ln r_2-ln r_1)$$
$$implies A=frac{V_2 - V_1}{ln r_2 - ln r_1} = frac{V_2 - V_1}{ln (r_2/r_1)}$$
Pluging this result into $(1)$ we get
$$begin{align}
V_1&=frac{V_2 - V_1}{ln (r_2/r_1)}ln r_1 + B \
implies B&=V_1-frac{V_2 - V_1}{ln (r_2/r_1)}ln r_1 \
&= frac{V_1 ln(r_2/r_1)-(V_2-V_1)ln r_1}{ln (r_2/r_1)} \
&=frac{V_1ln r_2-V_1ln r_1-V_2ln r_1+V_1ln r_1}{ln (r_2/r_1)} \
&=frac{V_1ln r_2 - V_2ln r_1}{ln (r_2/r_1)} \
end{align}$$
Inserting these into the equation $V(r)=Aln r+B$ gives
$$begin{align}
V(r) &= frac{V_2 - V_1}{ln (r_2/r_1)} ln r + frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} \
&=frac{V_2ln r - V_1ln r+V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} \
&=frac{V_2(ln r - ln r_1) - V_1(ln r - ln r_2)}{ln (r_2/r_1)} \
&=frac{V_2ln (r/r_1) - V_1ln (r/r_2)}{ln (r_2/r_1)}
end{align}$$
as desired.
answered Jan 23 at 20:54
greeliousgreelious
472112
$endgroup$
$begingroup$
That's great, thank you. Much appreciated, and makes sense.
$endgroup$
– freja
Jan 23 at 22:27
add a comment |
$begingroup$
Your equations are
$$V_1=Aln r_1+Btag1$$
$$V_2=Aln r_2+Btag2$$
Subtracting $(1)$ from $(2)$ we obtain
$$V_2-V_1=A(ln r_2-ln r_1)$$
$$implies A=frac{V_2 - V_1}{ln r_2 - ln r_1} = frac{V_2 - V_1}{ln (r_2/r_1)}$$
Pluging this result into $(1)$ we get
$$begin{align}
V_1&=frac{V_2 - V_1}{ln (r_2/r_1)}ln r_1 + B \
implies B&=V_1-frac{V_2 - V_1}{ln (r_2/r_1)}ln r_1 \
&= frac{V_1 ln(r_2/r_1)-(V_2-V_1)ln r_1}{ln (r_2/r_1)} \
&=frac{V_1ln r_2-V_1ln r_1-V_2ln r_1+V_1ln r_1}{ln (r_2/r_1)} \
&=frac{V_1ln r_2 - V_2ln r_1}{ln (r_2/r_1)} \
end{align}$$
Inserting these into the equation $V(r)=Aln r+B$ gives
$$begin{align}
V(r) &= frac{V_2 - V_1}{ln (r_2/r_1)} ln r + frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} \
&=frac{V_2ln r - V_1ln r+V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} \
&=frac{V_2(ln r - ln r_1) - V_1(ln r - ln r_2)}{ln (r_2/r_1)} \
&=frac{V_2ln (r/r_1) - V_1ln (r/r_2)}{ln (r_2/r_1)}
end{align}$$
as desired.
answered Jan 23 at 20:54
greeliousgreelious
472112
$endgroup$
$begingroup$
That's great, thank you. Much appreciated, and makes sense.
$endgroup$
– freja
Jan 23 at 22:27
add a comment |
$begingroup$
Your equations are
$$V_1=Aln r_1+Btag1$$
$$V_2=Aln r_2+Btag2$$
Subtracting $(1)$ from $(2)$ we obtain
$$V_2-V_1=A(ln r_2-ln r_1)$$
$$implies A=frac{V_2 - V_1}{ln r_2 - ln r_1} = frac{V_2 - V_1}{ln (r_2/r_1)}$$
Pluging this result into $(1)$ we get
$$begin{align}
V_1&=frac{V_2 - V_1}{ln (r_2/r_1)}ln r_1 + B \
implies B&=V_1-frac{V_2 - V_1}{ln (r_2/r_1)}ln r_1 \
&= frac{V_1 ln(r_2/r_1)-(V_2-V_1)ln r_1}{ln (r_2/r_1)} \
&=frac{V_1ln r_2-V_1ln r_1-V_2ln r_1+V_1ln r_1}{ln (r_2/r_1)} \
&=frac{V_1ln r_2 - V_2ln r_1}{ln (r_2/r_1)} \
end{align}$$
Inserting these into the equation $V(r)=Aln r+B$ gives
$$begin{align}
V(r) &= frac{V_2 - V_1}{ln (r_2/r_1)} ln r + frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} \
&=frac{V_2ln r - V_1ln r+V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} \
&=frac{V_2(ln r - ln r_1) - V_1(ln r - ln r_2)}{ln (r_2/r_1)} \
&=frac{V_2ln (r/r_1) - V_1ln (r/r_2)}{ln (r_2/r_1)}
end{align}$$
as desired.
answered Jan 23 at 20:54
greeliousgreelious
472112
$endgroup$
Your equations are
$$V_1=Aln r_1+Btag1$$
$$V_2=Aln r_2+Btag2$$
Subtracting $(1)$ from $(2)$ we obtain
$$V_2-V_1=A(ln r_2-ln r_1)$$
$$implies A=frac{V_2 - V_1}{ln r_2 - ln r_1} = frac{V_2 - V_1}{ln (r_2/r_1)}$$
Pluging this result into $(1)$ we get
$$begin{align}
V_1&=frac{V_2 - V_1}{ln (r_2/r_1)}ln r_1 + B \
implies B&=V_1-frac{V_2 - V_1}{ln (r_2/r_1)}ln r_1 \
&= frac{V_1 ln(r_2/r_1)-(V_2-V_1)ln r_1}{ln (r_2/r_1)} \
&=frac{V_1ln r_2-V_1ln r_1-V_2ln r_1+V_1ln r_1}{ln (r_2/r_1)} \
&=frac{V_1ln r_2 - V_2ln r_1}{ln (r_2/r_1)} \
end{align}$$
Inserting these into the equation $V(r)=Aln r+B$ gives
$$begin{align}
V(r) &= frac{V_2 - V_1}{ln (r_2/r_1)} ln r + frac{V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} \
&=frac{V_2ln r - V_1ln r+V_1 ln r_2 - V_2 ln r_1}{ln (r_2/r_1)} \
&=frac{V_2(ln r - ln r_1) - V_1(ln r - ln r_2)}{ln (r_2/r_1)} \
&=frac{V_2ln (r/r_1) - V_1ln (r/r_2)}{ln (r_2/r_1)}
end{align}$$
as desired.
answered Jan 23 at 20:54
greeliousgreelious
472112
answered Jan 23 at 20:54
greeliousgreelious
472112
answered Jan 23 at 20:54
greeliousgreelious
472112
answered Jan 23 at 20:54
greeliousgreelious
472112
472112
$begingroup$
That's great, thank you. Much appreciated, and makes sense.
$endgroup$
– freja
Jan 23 at 22:27
add a comment |
$begingroup$
That's great, thank you. Much appreciated, and makes sense.
$endgroup$
– freja
Jan 23 at 22:27
$begingroup$
That's great, thank you. Much appreciated, and makes sense.
$endgroup$
– freja
Jan 23 at 22:27
$begingroup$
That's great, thank you. Much appreciated, and makes sense.
$endgroup$
– freja
Jan 23 at 22:27
add a comment |
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$begingroup$
What is wrong with the answer in the book? It is a simple solution to a simple problem. Your approach seems to be unnecessarily complicated.
$endgroup$
– herb steinberg
Jan 23 at 20:43
$begingroup$
What do you mean? The last line is the book answer, I don't understand how they got there.
$endgroup$
– freja
Jan 23 at 22:23