Mental approximations of 1) log, 2) non-integer power
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I am currently preparing for a interview that is notorious for asking mental approximations. Two example questions came up: 1) $ln 514$ and 2) $3^{3.6}$.
What are some of the best ways to calculate these on the spot? For 1), my consideration was to use the change of base $ln 514 = log_2 514/log_2 e approx log_2 512/0.7 = 90/7$, but even that seems like a bit of a stretch. For 2), my approach was to use $3^{3.6}>3^3sqrt{3}approx 27times 1.7$, which once again seems like a stretch.
Does anyone have better/more efficient/creative ways of approximating these? Please share, thank you!
logarithms approximation mental-arithmetic
asked Dec 19 '18 at 13:58
user107224user107224
456314
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show 2 more comments
$begingroup$
I am currently preparing for a interview that is notorious for asking mental approximations. Two example questions came up: 1) $ln 514$ and 2) $3^{3.6}$.
What are some of the best ways to calculate these on the spot? For 1), my consideration was to use the change of base $ln 514 = log_2 514/log_2 e approx log_2 512/0.7 = 90/7$, but even that seems like a bit of a stretch. For 2), my approach was to use $3^{3.6}>3^3sqrt{3}approx 27times 1.7$, which once again seems like a stretch.
Does anyone have better/more efficient/creative ways of approximating these? Please share, thank you!
logarithms approximation mental-arithmetic
asked Dec 19 '18 at 13:58
user107224user107224
456314
$endgroup$
$begingroup$
for natural logarithm, I would use $ln 514 = ln 5.14 + 2 ln 10 approx ln 5 +2 cdot ln 10$. If you know that $ln 10 approx 2.3$, you can get pretty close approximation.
$endgroup$
– Vasya
Dec 19 '18 at 14:11
1
$begingroup$
I would go $ln 514 approx ln 512 = 9 ln 2 approx 6.3$ (using your estimate of $ln 2$), which is easier than dividing anything by 0.7
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– Max Freiburghaus
Dec 19 '18 at 14:14
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... and for the other one: $3^{3.6} = 3^3 times 3^{0.6} = 27 * 27^{1/5} approx 27*2=54$ because 27 is pretty close to 32 (considering that the gradient of $y=x^5$ is 80 at $x=2$)
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– Max Freiburghaus
Dec 19 '18 at 14:19
$begingroup$
What about working graphically ? With paper and pencil, you can sketch a reasonable $10^x$ curve between $0$ and $1$, and probably achieve accuracy better than $5%$.
$endgroup$
– Yves Daoust
Dec 19 '18 at 14:29
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$ln 2$ is about $0.7$, so $log_2 e approx 1/0.7$, and a good approximation to $ln 514$ would be $9 times 0.7 = 6.3$. (The actual value is about $6.242$.)
$endgroup$
– Michael Lugo
Dec 19 '18 at 15:36
|
show 2 more comments
$begingroup$
I am currently preparing for a interview that is notorious for asking mental approximations. Two example questions came up: 1) $ln 514$ and 2) $3^{3.6}$.
What are some of the best ways to calculate these on the spot? For 1), my consideration was to use the change of base $ln 514 = log_2 514/log_2 e approx log_2 512/0.7 = 90/7$, but even that seems like a bit of a stretch. For 2), my approach was to use $3^{3.6}>3^3sqrt{3}approx 27times 1.7$, which once again seems like a stretch.
Does anyone have better/more efficient/creative ways of approximating these? Please share, thank you!
logarithms approximation mental-arithmetic
asked Dec 19 '18 at 13:58
user107224user107224
456314
$endgroup$
I am currently preparing for a interview that is notorious for asking mental approximations. Two example questions came up: 1) $ln 514$ and 2) $3^{3.6}$.
What are some of the best ways to calculate these on the spot? For 1), my consideration was to use the change of base $ln 514 = log_2 514/log_2 e approx log_2 512/0.7 = 90/7$, but even that seems like a bit of a stretch. For 2), my approach was to use $3^{3.6}>3^3sqrt{3}approx 27times 1.7$, which once again seems like a stretch.
Does anyone have better/more efficient/creative ways of approximating these? Please share, thank you!
logarithms approximation mental-arithmetic
logarithms approximation mental-arithmetic
asked Dec 19 '18 at 13:58
user107224user107224
456314
asked Dec 19 '18 at 13:58
user107224user107224
456314
edited Dec 19 '18 at 14:15
user107224
asked Dec 19 '18 at 13:58
user107224user107224
456314
asked Dec 19 '18 at 13:58
user107224user107224
456314
asked Dec 19 '18 at 13:58
user107224user107224
456314
456314
$begingroup$
for natural logarithm, I would use $ln 514 = ln 5.14 + 2 ln 10 approx ln 5 +2 cdot ln 10$. If you know that $ln 10 approx 2.3$, you can get pretty close approximation.
$endgroup$
– Vasya
Dec 19 '18 at 14:11
1
$begingroup$
I would go $ln 514 approx ln 512 = 9 ln 2 approx 6.3$ (using your estimate of $ln 2$), which is easier than dividing anything by 0.7
$endgroup$
– Max Freiburghaus
Dec 19 '18 at 14:14
$begingroup$
... and for the other one: $3^{3.6} = 3^3 times 3^{0.6} = 27 * 27^{1/5} approx 27*2=54$ because 27 is pretty close to 32 (considering that the gradient of $y=x^5$ is 80 at $x=2$)
$endgroup$
– Max Freiburghaus
Dec 19 '18 at 14:19
$begingroup$
What about working graphically ? With paper and pencil, you can sketch a reasonable $10^x$ curve between $0$ and $1$, and probably achieve accuracy better than $5%$.
$endgroup$
– Yves Daoust
Dec 19 '18 at 14:29
$begingroup$
$ln 2$ is about $0.7$, so $log_2 e approx 1/0.7$, and a good approximation to $ln 514$ would be $9 times 0.7 = 6.3$. (The actual value is about $6.242$.)
$endgroup$
– Michael Lugo
Dec 19 '18 at 15:36
|
show 2 more comments
$begingroup$
for natural logarithm, I would use $ln 514 = ln 5.14 + 2 ln 10 approx ln 5 +2 cdot ln 10$. If you know that $ln 10 approx 2.3$, you can get pretty close approximation.
$endgroup$
– Vasya
Dec 19 '18 at 14:11
1
$begingroup$
I would go $ln 514 approx ln 512 = 9 ln 2 approx 6.3$ (using your estimate of $ln 2$), which is easier than dividing anything by 0.7
$endgroup$
– Max Freiburghaus
Dec 19 '18 at 14:14
$begingroup$
... and for the other one: $3^{3.6} = 3^3 times 3^{0.6} = 27 * 27^{1/5} approx 27*2=54$ because 27 is pretty close to 32 (considering that the gradient of $y=x^5$ is 80 at $x=2$)
$endgroup$
– Max Freiburghaus
Dec 19 '18 at 14:19
$begingroup$
What about working graphically ? With paper and pencil, you can sketch a reasonable $10^x$ curve between $0$ and $1$, and probably achieve accuracy better than $5%$.
$endgroup$
– Yves Daoust
Dec 19 '18 at 14:29
$begingroup$
$ln 2$ is about $0.7$, so $log_2 e approx 1/0.7$, and a good approximation to $ln 514$ would be $9 times 0.7 = 6.3$. (The actual value is about $6.242$.)
$endgroup$
– Michael Lugo
Dec 19 '18 at 15:36
$begingroup$
for natural logarithm, I would use $ln 514 = ln 5.14 + 2 ln 10 approx ln 5 +2 cdot ln 10$. If you know that $ln 10 approx 2.3$, you can get pretty close approximation.
$endgroup$
– Vasya
Dec 19 '18 at 14:11
$begingroup$
for natural logarithm, I would use $ln 514 = ln 5.14 + 2 ln 10 approx ln 5 +2 cdot ln 10$. If you know that $ln 10 approx 2.3$, you can get pretty close approximation.
$endgroup$
– Vasya
Dec 19 '18 at 14:11
1
1
$begingroup$
I would go $ln 514 approx ln 512 = 9 ln 2 approx 6.3$ (using your estimate of $ln 2$), which is easier than dividing anything by 0.7
$endgroup$
– Max Freiburghaus
Dec 19 '18 at 14:14
$begingroup$
I would go $ln 514 approx ln 512 = 9 ln 2 approx 6.3$ (using your estimate of $ln 2$), which is easier than dividing anything by 0.7
$endgroup$
– Max Freiburghaus
Dec 19 '18 at 14:14
$begingroup$
... and for the other one: $3^{3.6} = 3^3 times 3^{0.6} = 27 * 27^{1/5} approx 27*2=54$ because 27 is pretty close to 32 (considering that the gradient of $y=x^5$ is 80 at $x=2$)
$endgroup$
– Max Freiburghaus
Dec 19 '18 at 14:19
$begingroup$
... and for the other one: $3^{3.6} = 3^3 times 3^{0.6} = 27 * 27^{1/5} approx 27*2=54$ because 27 is pretty close to 32 (considering that the gradient of $y=x^5$ is 80 at $x=2$)
$endgroup$
– Max Freiburghaus
Dec 19 '18 at 14:19
$begingroup$
What about working graphically ? With paper and pencil, you can sketch a reasonable $10^x$ curve between $0$ and $1$, and probably achieve accuracy better than $5%$.
$endgroup$
– Yves Daoust
Dec 19 '18 at 14:29
$begingroup$
What about working graphically ? With paper and pencil, you can sketch a reasonable $10^x$ curve between $0$ and $1$, and probably achieve accuracy better than $5%$.
$endgroup$
– Yves Daoust
Dec 19 '18 at 14:29
$begingroup$
$ln 2$ is about $0.7$, so $log_2 e approx 1/0.7$, and a good approximation to $ln 514$ would be $9 times 0.7 = 6.3$. (The actual value is about $6.242$.)
$endgroup$
– Michael Lugo
Dec 19 '18 at 15:36
$begingroup$
$ln 2$ is about $0.7$, so $log_2 e approx 1/0.7$, and a good approximation to $ln 514$ would be $9 times 0.7 = 6.3$. (The actual value is about $6.242$.)
$endgroup$
– Michael Lugo
Dec 19 '18 at 15:36
|
show 2 more comments
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$begingroup$
for natural logarithm, I would use $ln 514 = ln 5.14 + 2 ln 10 approx ln 5 +2 cdot ln 10$. If you know that $ln 10 approx 2.3$, you can get pretty close approximation.
$endgroup$
– Vasya
Dec 19 '18 at 14:11
1
$begingroup$
I would go $ln 514 approx ln 512 = 9 ln 2 approx 6.3$ (using your estimate of $ln 2$), which is easier than dividing anything by 0.7
$endgroup$
– Max Freiburghaus
Dec 19 '18 at 14:14
$begingroup$
... and for the other one: $3^{3.6} = 3^3 times 3^{0.6} = 27 * 27^{1/5} approx 27*2=54$ because 27 is pretty close to 32 (considering that the gradient of $y=x^5$ is 80 at $x=2$)
$endgroup$
– Max Freiburghaus
Dec 19 '18 at 14:19
$begingroup$
What about working graphically ? With paper and pencil, you can sketch a reasonable $10^x$ curve between $0$ and $1$, and probably achieve accuracy better than $5%$.
$endgroup$
– Yves Daoust
Dec 19 '18 at 14:29
$begingroup$
$ln 2$ is about $0.7$, so $log_2 e approx 1/0.7$, and a good approximation to $ln 514$ would be $9 times 0.7 = 6.3$. (The actual value is about $6.242$.)
$endgroup$
– Michael Lugo
Dec 19 '18 at 15:36