Why is it that $lnBig(1+frac{x}{y}Big)=ln(1+exp(ln x - ln y))$?
Is the relation $lnBig(1+frac{x}{y}Big)=ln(1+exp(ln x - ln y))$ an approximation?
If so, how can I derive this relation?
algebra-precalculus logarithms exponential-function
add a comment |
Is the relation $lnBig(1+frac{x}{y}Big)=ln(1+exp(ln x - ln y))$ an approximation?
If so, how can I derive this relation?
algebra-precalculus logarithms exponential-function
1
Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
– Alecos Papadopoulos
Nov 29 at 19:33
add a comment |
Is the relation $lnBig(1+frac{x}{y}Big)=ln(1+exp(ln x - ln y))$ an approximation?
If so, how can I derive this relation?
algebra-precalculus logarithms exponential-function
Is the relation $lnBig(1+frac{x}{y}Big)=ln(1+exp(ln x - ln y))$ an approximation?
If so, how can I derive this relation?
algebra-precalculus logarithms exponential-function
algebra-precalculus logarithms exponential-function
edited Nov 29 at 19:29
Foobaz John
20.7k41250
20.7k41250
asked Nov 29 at 19:27
Jeremy Kwak
92
92
1
Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
– Alecos Papadopoulos
Nov 29 at 19:33
add a comment |
1
Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
– Alecos Papadopoulos
Nov 29 at 19:33
1
1
Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
– Alecos Papadopoulos
Nov 29 at 19:33
Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
– Alecos Papadopoulos
Nov 29 at 19:33
add a comment |
3 Answers
3
active
oldest
votes
$$ ln(x) - ln(y) = lnleft(frac{x}{y} right) \
expleft(lnleft(frac{x}{y}right)right) = frac{x}{y} \
implies lnleft(1 + frac{x}{y}right)
$$
3
Note that you can useexp
andln
to format it better.
– TheSimpliFire
Nov 29 at 19:37
add a comment |
It’s not an approximation; it’s exact. Since the natural logarithm and $e^x$ are inverse functions, we know that, since $f^{-1}(f(x)=x$, $$e^{mathrm{ln} x}=x$$
Replacing $x$ with $frac{x}{y}$, we have $$e^{ln(frac{x}{y})}=frac{x}{y}$$
So, plugging this into the original equation $$lnleft( 1 + frac{x}{y}right)=lnleft(1+e^{lnleft(frac{x}{y}right)}right)$$
We know that a property of all logarithms is that $$log_w(frac{x}{y})=log_w(x)-log_w(y)$$
So, substituting we have $$lnleft(1+e^{lnleft(frac{x}{y}right)}right)= lnleft(1+e^{ln(x)-ln(y)}right)= lnleft(1+exp(ln(x)-ln(y))right)$$
add a comment |
Note that
$$log a-log b = logfrac{a}{b}$$
Applying that here, you get
$$ln big(1+e^{ln x-ln y}big) = ln big(1+e^{ln frac{x}{y}}big)$$
Also, as inverses, $e^{ln a} = a$ (and $ln e^a = a$), meaning the expression simplifies to
$$ln bigg(1+frac{x}{y}bigg)$$
Also, to point out, this isn’t an approximation. The two expressions are exactly equivalent.
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
$$ ln(x) - ln(y) = lnleft(frac{x}{y} right) \
expleft(lnleft(frac{x}{y}right)right) = frac{x}{y} \
implies lnleft(1 + frac{x}{y}right)
$$
3
Note that you can useexp
andln
to format it better.
– TheSimpliFire
Nov 29 at 19:37
add a comment |
$$ ln(x) - ln(y) = lnleft(frac{x}{y} right) \
expleft(lnleft(frac{x}{y}right)right) = frac{x}{y} \
implies lnleft(1 + frac{x}{y}right)
$$
3
Note that you can useexp
andln
to format it better.
– TheSimpliFire
Nov 29 at 19:37
add a comment |
$$ ln(x) - ln(y) = lnleft(frac{x}{y} right) \
expleft(lnleft(frac{x}{y}right)right) = frac{x}{y} \
implies lnleft(1 + frac{x}{y}right)
$$
$$ ln(x) - ln(y) = lnleft(frac{x}{y} right) \
expleft(lnleft(frac{x}{y}right)right) = frac{x}{y} \
implies lnleft(1 + frac{x}{y}right)
$$
edited Nov 29 at 20:12
Eevee Trainer
3,390225
3,390225
answered Nov 29 at 19:34
Филип Димитровски
314
314
3
Note that you can useexp
andln
to format it better.
– TheSimpliFire
Nov 29 at 19:37
add a comment |
3
Note that you can useexp
andln
to format it better.
– TheSimpliFire
Nov 29 at 19:37
3
3
Note that you can use
exp
and ln
to format it better.– TheSimpliFire
Nov 29 at 19:37
Note that you can use
exp
and ln
to format it better.– TheSimpliFire
Nov 29 at 19:37
add a comment |
It’s not an approximation; it’s exact. Since the natural logarithm and $e^x$ are inverse functions, we know that, since $f^{-1}(f(x)=x$, $$e^{mathrm{ln} x}=x$$
Replacing $x$ with $frac{x}{y}$, we have $$e^{ln(frac{x}{y})}=frac{x}{y}$$
So, plugging this into the original equation $$lnleft( 1 + frac{x}{y}right)=lnleft(1+e^{lnleft(frac{x}{y}right)}right)$$
We know that a property of all logarithms is that $$log_w(frac{x}{y})=log_w(x)-log_w(y)$$
So, substituting we have $$lnleft(1+e^{lnleft(frac{x}{y}right)}right)= lnleft(1+e^{ln(x)-ln(y)}right)= lnleft(1+exp(ln(x)-ln(y))right)$$
add a comment |
It’s not an approximation; it’s exact. Since the natural logarithm and $e^x$ are inverse functions, we know that, since $f^{-1}(f(x)=x$, $$e^{mathrm{ln} x}=x$$
Replacing $x$ with $frac{x}{y}$, we have $$e^{ln(frac{x}{y})}=frac{x}{y}$$
So, plugging this into the original equation $$lnleft( 1 + frac{x}{y}right)=lnleft(1+e^{lnleft(frac{x}{y}right)}right)$$
We know that a property of all logarithms is that $$log_w(frac{x}{y})=log_w(x)-log_w(y)$$
So, substituting we have $$lnleft(1+e^{lnleft(frac{x}{y}right)}right)= lnleft(1+e^{ln(x)-ln(y)}right)= lnleft(1+exp(ln(x)-ln(y))right)$$
add a comment |
It’s not an approximation; it’s exact. Since the natural logarithm and $e^x$ are inverse functions, we know that, since $f^{-1}(f(x)=x$, $$e^{mathrm{ln} x}=x$$
Replacing $x$ with $frac{x}{y}$, we have $$e^{ln(frac{x}{y})}=frac{x}{y}$$
So, plugging this into the original equation $$lnleft( 1 + frac{x}{y}right)=lnleft(1+e^{lnleft(frac{x}{y}right)}right)$$
We know that a property of all logarithms is that $$log_w(frac{x}{y})=log_w(x)-log_w(y)$$
So, substituting we have $$lnleft(1+e^{lnleft(frac{x}{y}right)}right)= lnleft(1+e^{ln(x)-ln(y)}right)= lnleft(1+exp(ln(x)-ln(y))right)$$
It’s not an approximation; it’s exact. Since the natural logarithm and $e^x$ are inverse functions, we know that, since $f^{-1}(f(x)=x$, $$e^{mathrm{ln} x}=x$$
Replacing $x$ with $frac{x}{y}$, we have $$e^{ln(frac{x}{y})}=frac{x}{y}$$
So, plugging this into the original equation $$lnleft( 1 + frac{x}{y}right)=lnleft(1+e^{lnleft(frac{x}{y}right)}right)$$
We know that a property of all logarithms is that $$log_w(frac{x}{y})=log_w(x)-log_w(y)$$
So, substituting we have $$lnleft(1+e^{lnleft(frac{x}{y}right)}right)= lnleft(1+e^{ln(x)-ln(y)}right)= lnleft(1+exp(ln(x)-ln(y))right)$$
edited Nov 29 at 19:46
answered Nov 29 at 19:43
Tesseract
21615
21615
add a comment |
add a comment |
Note that
$$log a-log b = logfrac{a}{b}$$
Applying that here, you get
$$ln big(1+e^{ln x-ln y}big) = ln big(1+e^{ln frac{x}{y}}big)$$
Also, as inverses, $e^{ln a} = a$ (and $ln e^a = a$), meaning the expression simplifies to
$$ln bigg(1+frac{x}{y}bigg)$$
Also, to point out, this isn’t an approximation. The two expressions are exactly equivalent.
add a comment |
Note that
$$log a-log b = logfrac{a}{b}$$
Applying that here, you get
$$ln big(1+e^{ln x-ln y}big) = ln big(1+e^{ln frac{x}{y}}big)$$
Also, as inverses, $e^{ln a} = a$ (and $ln e^a = a$), meaning the expression simplifies to
$$ln bigg(1+frac{x}{y}bigg)$$
Also, to point out, this isn’t an approximation. The two expressions are exactly equivalent.
add a comment |
Note that
$$log a-log b = logfrac{a}{b}$$
Applying that here, you get
$$ln big(1+e^{ln x-ln y}big) = ln big(1+e^{ln frac{x}{y}}big)$$
Also, as inverses, $e^{ln a} = a$ (and $ln e^a = a$), meaning the expression simplifies to
$$ln bigg(1+frac{x}{y}bigg)$$
Also, to point out, this isn’t an approximation. The two expressions are exactly equivalent.
Note that
$$log a-log b = logfrac{a}{b}$$
Applying that here, you get
$$ln big(1+e^{ln x-ln y}big) = ln big(1+e^{ln frac{x}{y}}big)$$
Also, as inverses, $e^{ln a} = a$ (and $ln e^a = a$), meaning the expression simplifies to
$$ln bigg(1+frac{x}{y}bigg)$$
Also, to point out, this isn’t an approximation. The two expressions are exactly equivalent.
answered Nov 29 at 20:07
KM101
3,958417
3,958417
add a comment |
add a comment |
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1
Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
– Alecos Papadopoulos
Nov 29 at 19:33