Simplify $log(α+ b(x- t)+ k)$
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I am trying to simplify this natural log expression the best that I can, but I am unsure what to do in order to separate $b(x-t)$. Would it be $log(b) + log (x/t)$? so would the whole thing be:
$$ log (alpha + b) + log (x/t) + log k quad?$$
Thank you!
logarithms
edited Dec 8 '18 at 16:01
Dando18
4,67741235
asked Dec 8 '18 at 15:43
Beary1213Beary1213
1
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add a comment |
$begingroup$
I am trying to simplify this natural log expression the best that I can, but I am unsure what to do in order to separate $b(x-t)$. Would it be $log(b) + log (x/t)$? so would the whole thing be:
$$ log (alpha + b) + log (x/t) + log k quad?$$
Thank you!
logarithms
edited Dec 8 '18 at 16:01
Dando18
4,67741235
asked Dec 8 '18 at 15:43
Beary1213Beary1213
1
$endgroup$
$begingroup$
Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
$endgroup$
– KM101
Dec 8 '18 at 15:52
add a comment |
$begingroup$
I am trying to simplify this natural log expression the best that I can, but I am unsure what to do in order to separate $b(x-t)$. Would it be $log(b) + log (x/t)$? so would the whole thing be:
$$ log (alpha + b) + log (x/t) + log k quad?$$
Thank you!
logarithms
edited Dec 8 '18 at 16:01
Dando18
4,67741235
asked Dec 8 '18 at 15:43
Beary1213Beary1213
1
$endgroup$
I am trying to simplify this natural log expression the best that I can, but I am unsure what to do in order to separate $b(x-t)$. Would it be $log(b) + log (x/t)$? so would the whole thing be:
$$ log (alpha + b) + log (x/t) + log k quad?$$
Thank you!
logarithms
logarithms
edited Dec 8 '18 at 16:01
Dando18
4,67741235
asked Dec 8 '18 at 15:43
Beary1213Beary1213
1
edited Dec 8 '18 at 16:01
Dando18
4,67741235
asked Dec 8 '18 at 15:43
Beary1213Beary1213
1
edited Dec 8 '18 at 16:01
Dando18
4,67741235
edited Dec 8 '18 at 16:01
Dando18
4,67741235
edited Dec 8 '18 at 16:01
Dando18
4,67741235
4,67741235
asked Dec 8 '18 at 15:43
Beary1213Beary1213
1
asked Dec 8 '18 at 15:43
Beary1213Beary1213
1
asked Dec 8 '18 at 15:43
Beary1213Beary1213
1
1
$begingroup$
Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
$endgroup$
– KM101
Dec 8 '18 at 15:52
add a comment |
$begingroup$
Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
$endgroup$
– KM101
Dec 8 '18 at 15:52
$begingroup$
Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
$endgroup$
– KM101
Dec 8 '18 at 15:52
$begingroup$
Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
$endgroup$
– KM101
Dec 8 '18 at 15:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No. You can't separate $log(A+B)$ into two logarithms, no matter what $A$ and $B$ are. The only things you can separate are $log(AB)=log(A)+log(B)$, $log_B(A)=frac{log(A)}{log(B)}$ and $log(A^B)=Blog(A)$, and their inverse operations
answered Dec 8 '18 at 15:50
MoKo19MoKo19
1914
$endgroup$
add a comment |
$begingroup$
You can only write $$log(alpha+bx-bt+k)$$ it holds only
$$log(ab)=log(a)+log(b)$$ and $$logfrac{a}{b}=log(a)-log(b)$$ for positive variables $a,b$
answered Dec 8 '18 at 15:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
No. You can't separate $log(A+B)$ into two logarithms, no matter what $A$ and $B$ are. The only things you can separate are $log(AB)=log(A)+log(B)$, $log_B(A)=frac{log(A)}{log(B)}$ and $log(A^B)=Blog(A)$, and their inverse operations
answered Dec 8 '18 at 15:50
MoKo19MoKo19
1914
$endgroup$
add a comment |
$begingroup$
No. You can't separate $log(A+B)$ into two logarithms, no matter what $A$ and $B$ are. The only things you can separate are $log(AB)=log(A)+log(B)$, $log_B(A)=frac{log(A)}{log(B)}$ and $log(A^B)=Blog(A)$, and their inverse operations
answered Dec 8 '18 at 15:50
MoKo19MoKo19
1914
$endgroup$
add a comment |
$begingroup$
No. You can't separate $log(A+B)$ into two logarithms, no matter what $A$ and $B$ are. The only things you can separate are $log(AB)=log(A)+log(B)$, $log_B(A)=frac{log(A)}{log(B)}$ and $log(A^B)=Blog(A)$, and their inverse operations
answered Dec 8 '18 at 15:50
MoKo19MoKo19
1914
$endgroup$
No. You can't separate $log(A+B)$ into two logarithms, no matter what $A$ and $B$ are. The only things you can separate are $log(AB)=log(A)+log(B)$, $log_B(A)=frac{log(A)}{log(B)}$ and $log(A^B)=Blog(A)$, and their inverse operations
answered Dec 8 '18 at 15:50
MoKo19MoKo19
1914
answered Dec 8 '18 at 15:50
MoKo19MoKo19
1914
answered Dec 8 '18 at 15:50
MoKo19MoKo19
1914
answered Dec 8 '18 at 15:50
MoKo19MoKo19
1914
1914
add a comment |
add a comment |
$begingroup$
You can only write $$log(alpha+bx-bt+k)$$ it holds only
$$log(ab)=log(a)+log(b)$$ and $$logfrac{a}{b}=log(a)-log(b)$$ for positive variables $a,b$
answered Dec 8 '18 at 15:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
$endgroup$
add a comment |
$begingroup$
You can only write $$log(alpha+bx-bt+k)$$ it holds only
$$log(ab)=log(a)+log(b)$$ and $$logfrac{a}{b}=log(a)-log(b)$$ for positive variables $a,b$
answered Dec 8 '18 at 15:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
$endgroup$
add a comment |
$begingroup$
You can only write $$log(alpha+bx-bt+k)$$ it holds only
$$log(ab)=log(a)+log(b)$$ and $$logfrac{a}{b}=log(a)-log(b)$$ for positive variables $a,b$
answered Dec 8 '18 at 15:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
$endgroup$
You can only write $$log(alpha+bx-bt+k)$$ it holds only
$$log(ab)=log(a)+log(b)$$ and $$logfrac{a}{b}=log(a)-log(b)$$ for positive variables $a,b$
answered Dec 8 '18 at 15:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
answered Dec 8 '18 at 15:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
answered Dec 8 '18 at 15:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
answered Dec 8 '18 at 15:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
74.6k42865
add a comment |
add a comment |
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$begingroup$
Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
$endgroup$
– KM101
Dec 8 '18 at 15:52