logarithmic transformation from exponential to linear equation
$begingroup$
How to convert this exponential equation to linear equation.
$Y =exp(17.9348)cdot x^{-2.705}$
what I did is:
$Y =log(17.9348)-2.705log(x).$
I am confused with this one:
$Y=17.9348-2.705log(x).$
Which one is correct transformation?
Thanks in advance!
logarithms linear-transformations transformation
edited Jan 11 at 16:18
Adrian Keister
5,27371933
asked Jan 3 at 15:39
user3063user3063
1143
$endgroup$
add a comment |
$begingroup$
How to convert this exponential equation to linear equation.
$Y =exp(17.9348)cdot x^{-2.705}$
what I did is:
$Y =log(17.9348)-2.705log(x).$
I am confused with this one:
$Y=17.9348-2.705log(x).$
Which one is correct transformation?
Thanks in advance!
logarithms linear-transformations transformation
edited Jan 11 at 16:18
Adrian Keister
5,27371933
asked Jan 3 at 15:39
user3063user3063
1143
$endgroup$
add a comment |
$begingroup$
How to convert this exponential equation to linear equation.
$Y =exp(17.9348)cdot x^{-2.705}$
what I did is:
$Y =log(17.9348)-2.705log(x).$
I am confused with this one:
$Y=17.9348-2.705log(x).$
Which one is correct transformation?
Thanks in advance!
logarithms linear-transformations transformation
edited Jan 11 at 16:18
Adrian Keister
5,27371933
asked Jan 3 at 15:39
user3063user3063
1143
$endgroup$
How to convert this exponential equation to linear equation.
$Y =exp(17.9348)cdot x^{-2.705}$
what I did is:
$Y =log(17.9348)-2.705log(x).$
I am confused with this one:
$Y=17.9348-2.705log(x).$
Which one is correct transformation?
Thanks in advance!
logarithms linear-transformations transformation
logarithms linear-transformations transformation
edited Jan 11 at 16:18
Adrian Keister
5,27371933
asked Jan 3 at 15:39
user3063user3063
1143
edited Jan 11 at 16:18
Adrian Keister
5,27371933
asked Jan 3 at 15:39
user3063user3063
1143
edited Jan 11 at 16:18
Adrian Keister
5,27371933
edited Jan 11 at 16:18
Adrian Keister
5,27371933
edited Jan 11 at 16:18
Adrian Keister
5,27371933
5,27371933
asked Jan 3 at 15:39
user3063user3063
1143
asked Jan 3 at 15:39
user3063user3063
1143
asked Jan 3 at 15:39
user3063user3063
1143
1143
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer is neither! You need to take the logarithm of both sides:
begin{align*}
y&=e^{17.9348},x^{-2.705} \
ln(y)&=lnleft(e^{17.9348},x^{-2.705}right) \
ln(y)&=lnleft(e^{17.9348}right)+lnleft(x^{-2.705}right) \
ln(y)&=17.9348-2.705ln(x).
end{align*}
answered Jan 3 at 15:46
Adrian KeisterAdrian Keister
5,27371933
$endgroup$
$begingroup$
ok, thank you so much!
$endgroup$
– user3063
Jan 3 at 15:53
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The answer is neither! You need to take the logarithm of both sides:
begin{align*}
y&=e^{17.9348},x^{-2.705} \
ln(y)&=lnleft(e^{17.9348},x^{-2.705}right) \
ln(y)&=lnleft(e^{17.9348}right)+lnleft(x^{-2.705}right) \
ln(y)&=17.9348-2.705ln(x).
end{align*}
answered Jan 3 at 15:46
Adrian KeisterAdrian Keister
5,27371933
$endgroup$
$begingroup$
ok, thank you so much!
$endgroup$
– user3063
Jan 3 at 15:53
add a comment |
$begingroup$
The answer is neither! You need to take the logarithm of both sides:
begin{align*}
y&=e^{17.9348},x^{-2.705} \
ln(y)&=lnleft(e^{17.9348},x^{-2.705}right) \
ln(y)&=lnleft(e^{17.9348}right)+lnleft(x^{-2.705}right) \
ln(y)&=17.9348-2.705ln(x).
end{align*}
answered Jan 3 at 15:46
Adrian KeisterAdrian Keister
5,27371933
$endgroup$
$begingroup$
ok, thank you so much!
$endgroup$
– user3063
Jan 3 at 15:53
add a comment |
$begingroup$
The answer is neither! You need to take the logarithm of both sides:
begin{align*}
y&=e^{17.9348},x^{-2.705} \
ln(y)&=lnleft(e^{17.9348},x^{-2.705}right) \
ln(y)&=lnleft(e^{17.9348}right)+lnleft(x^{-2.705}right) \
ln(y)&=17.9348-2.705ln(x).
end{align*}
answered Jan 3 at 15:46
Adrian KeisterAdrian Keister
5,27371933
$endgroup$
The answer is neither! You need to take the logarithm of both sides:
begin{align*}
y&=e^{17.9348},x^{-2.705} \
ln(y)&=lnleft(e^{17.9348},x^{-2.705}right) \
ln(y)&=lnleft(e^{17.9348}right)+lnleft(x^{-2.705}right) \
ln(y)&=17.9348-2.705ln(x).
end{align*}
answered Jan 3 at 15:46
Adrian KeisterAdrian Keister
5,27371933
answered Jan 3 at 15:46
Adrian KeisterAdrian Keister
5,27371933
answered Jan 3 at 15:46
Adrian KeisterAdrian Keister
5,27371933
answered Jan 3 at 15:46
Adrian KeisterAdrian Keister
5,27371933
5,27371933
$begingroup$
ok, thank you so much!
$endgroup$
– user3063
Jan 3 at 15:53
add a comment |
$begingroup$
ok, thank you so much!
$endgroup$
– user3063
Jan 3 at 15:53
$begingroup$
ok, thank you so much!
$endgroup$
– user3063
Jan 3 at 15:53
$begingroup$
ok, thank you so much!
$endgroup$
– user3063
Jan 3 at 15:53
add a comment |
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