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Sort those 3 logarithmic values without using calculator
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I found this problem interesting, namely we are given three values: $$log_1/327, log_1/54, log_1/25$$
We want to sort those values without using calculator. So I decided to work on them a little bit before comparing anything. Here is what I got:
First of all: $log_1/327 = log_1/33^3 = log_1/3(frac13)^-3 = -3\ log_1/54 = log_1/52^2 = 2 cdot log_1/52 = 2 cdot log_5^-12 = -2 cdot log_52$
Since $log_52 < 1, textthen: -2cdot log_52 > -2 $
And the third value: $log_1/25 = log_2^-15 = -log_25$
Since $2 < log_25 < 3, textthen: -2 > -log_25 > -3 $
This shows us that the first values is the smallest, then the third one and finally the second one: $$log_1/327, log_1/25, log_1/54$$
Is my approach correct or have I made some mistakes, can we get more accurate values for those logarithms without using calculator.
logarithms
asked 5 hours ago
someone123123
382214
add a comment |Â
up vote
3
down vote
favorite
I found this problem interesting, namely we are given three values: $$log_1/327, log_1/54, log_1/25$$
We want to sort those values without using calculator. So I decided to work on them a little bit before comparing anything. Here is what I got:
First of all: $log_1/327 = log_1/33^3 = log_1/3(frac13)^-3 = -3\ log_1/54 = log_1/52^2 = 2 cdot log_1/52 = 2 cdot log_5^-12 = -2 cdot log_52$
Since $log_52 < 1, textthen: -2cdot log_52 > -2 $
And the third value: $log_1/25 = log_2^-15 = -log_25$
Since $2 < log_25 < 3, textthen: -2 > -log_25 > -3 $
This shows us that the first values is the smallest, then the third one and finally the second one: $$log_1/327, log_1/25, log_1/54$$
Is my approach correct or have I made some mistakes, can we get more accurate values for those logarithms without using calculator.
logarithms
asked 5 hours ago
someone123123
382214
Quite correct, though you can make it simpler.
– Yves Daoust
4 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I found this problem interesting, namely we are given three values: $$log_1/327, log_1/54, log_1/25$$
We want to sort those values without using calculator. So I decided to work on them a little bit before comparing anything. Here is what I got:
First of all: $log_1/327 = log_1/33^3 = log_1/3(frac13)^-3 = -3\ log_1/54 = log_1/52^2 = 2 cdot log_1/52 = 2 cdot log_5^-12 = -2 cdot log_52$
Since $log_52 < 1, textthen: -2cdot log_52 > -2 $
And the third value: $log_1/25 = log_2^-15 = -log_25$
Since $2 < log_25 < 3, textthen: -2 > -log_25 > -3 $
This shows us that the first values is the smallest, then the third one and finally the second one: $$log_1/327, log_1/25, log_1/54$$
Is my approach correct or have I made some mistakes, can we get more accurate values for those logarithms without using calculator.
logarithms
asked 5 hours ago
someone123123
382214
I found this problem interesting, namely we are given three values: $$log_1/327, log_1/54, log_1/25$$
We want to sort those values without using calculator. So I decided to work on them a little bit before comparing anything. Here is what I got:
First of all: $log_1/327 = log_1/33^3 = log_1/3(frac13)^-3 = -3\ log_1/54 = log_1/52^2 = 2 cdot log_1/52 = 2 cdot log_5^-12 = -2 cdot log_52$
Since $log_52 < 1, textthen: -2cdot log_52 > -2 $
And the third value: $log_1/25 = log_2^-15 = -log_25$
Since $2 < log_25 < 3, textthen: -2 > -log_25 > -3 $
This shows us that the first values is the smallest, then the third one and finally the second one: $$log_1/327, log_1/25, log_1/54$$
Is my approach correct or have I made some mistakes, can we get more accurate values for those logarithms without using calculator.
logarithms
logarithms
asked 5 hours ago
someone123123
382214
asked 5 hours ago
someone123123
382214
asked 5 hours ago
someone123123
382214
asked 5 hours ago
someone123123
382214
asked 5 hours ago
someone123123
382214
382214
Quite correct, though you can make it simpler.
– Yves Daoust
4 hours ago
add a comment |Â
Quite correct, though you can make it simpler.
– Yves Daoust
4 hours ago
Quite correct, though you can make it simpler.
– Yves Daoust
4 hours ago
Quite correct, though you can make it simpler.
– Yves Daoust
4 hours ago
add a comment |Â
2 Answers
2
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up vote
2
down vote
As an alternative and to check recall that
$$log_a b=fraclog_c alog_c b$$
that is
$$log_frac1327=fraclog_2 27log_2frac13=-fraclog_2 3^3log_2 3=-3$$
$$log_frac154=fraclog_2 4log_2 frac15=-fraclog_2 2^2log_2 5=-frac2log_2 5$$
$$log_frac125=fraclog_2 5log_2 frac12=-fraclog 5log_2 2=-log_2 5$$
therefore as you stated
$$log_frac1327<log_frac125<log_frac154$$
answered 5 hours ago
gimusi
73k73889
add a comment |Â
up vote
2
down vote
For clarity, we take the inverses of the bases, which will reverse the order ($log_1/xy=-log_x y$).
We have
$$log_327=3,$$
$$log_54<1,$$
because $4<5^1$ and
$$1<log_25<3$$
because $2^1<5<2^3$.
The rest is yours.
Short answer:
$$log_54<1<log_25<3=log_327$$
and
$$log_1/54>log_1/25>log_1/327.$$
answered 4 hours ago
Yves Daoust
114k666209
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
As an alternative and to check recall that
$$log_a b=fraclog_c alog_c b$$
that is
$$log_frac1327=fraclog_2 27log_2frac13=-fraclog_2 3^3log_2 3=-3$$
$$log_frac154=fraclog_2 4log_2 frac15=-fraclog_2 2^2log_2 5=-frac2log_2 5$$
$$log_frac125=fraclog_2 5log_2 frac12=-fraclog 5log_2 2=-log_2 5$$
therefore as you stated
$$log_frac1327<log_frac125<log_frac154$$
answered 5 hours ago
gimusi
73k73889
add a comment |Â
up vote
2
down vote
As an alternative and to check recall that
$$log_a b=fraclog_c alog_c b$$
that is
$$log_frac1327=fraclog_2 27log_2frac13=-fraclog_2 3^3log_2 3=-3$$
$$log_frac154=fraclog_2 4log_2 frac15=-fraclog_2 2^2log_2 5=-frac2log_2 5$$
$$log_frac125=fraclog_2 5log_2 frac12=-fraclog 5log_2 2=-log_2 5$$
therefore as you stated
$$log_frac1327<log_frac125<log_frac154$$
answered 5 hours ago
gimusi
73k73889
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As an alternative and to check recall that
$$log_a b=fraclog_c alog_c b$$
that is
$$log_frac1327=fraclog_2 27log_2frac13=-fraclog_2 3^3log_2 3=-3$$
$$log_frac154=fraclog_2 4log_2 frac15=-fraclog_2 2^2log_2 5=-frac2log_2 5$$
$$log_frac125=fraclog_2 5log_2 frac12=-fraclog 5log_2 2=-log_2 5$$
therefore as you stated
$$log_frac1327<log_frac125<log_frac154$$
answered 5 hours ago
gimusi
73k73889
As an alternative and to check recall that
$$log_a b=fraclog_c alog_c b$$
that is
$$log_frac1327=fraclog_2 27log_2frac13=-fraclog_2 3^3log_2 3=-3$$
$$log_frac154=fraclog_2 4log_2 frac15=-fraclog_2 2^2log_2 5=-frac2log_2 5$$
$$log_frac125=fraclog_2 5log_2 frac12=-fraclog 5log_2 2=-log_2 5$$
therefore as you stated
$$log_frac1327<log_frac125<log_frac154$$
answered 5 hours ago
gimusi
73k73889
answered 5 hours ago
gimusi
73k73889
answered 5 hours ago
gimusi
73k73889
answered 5 hours ago
gimusi
73k73889
73k73889
add a comment |Â
add a comment |Â
up vote
2
down vote
For clarity, we take the inverses of the bases, which will reverse the order ($log_1/xy=-log_x y$).
We have
$$log_327=3,$$
$$log_54<1,$$
because $4<5^1$ and
$$1<log_25<3$$
because $2^1<5<2^3$.
The rest is yours.
Short answer:
$$log_54<1<log_25<3=log_327$$
and
$$log_1/54>log_1/25>log_1/327.$$
answered 4 hours ago
Yves Daoust
114k666209
add a comment |Â
up vote
2
down vote
For clarity, we take the inverses of the bases, which will reverse the order ($log_1/xy=-log_x y$).
We have
$$log_327=3,$$
$$log_54<1,$$
because $4<5^1$ and
$$1<log_25<3$$
because $2^1<5<2^3$.
The rest is yours.
Short answer:
$$log_54<1<log_25<3=log_327$$
and
$$log_1/54>log_1/25>log_1/327.$$
answered 4 hours ago
Yves Daoust
114k666209
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For clarity, we take the inverses of the bases, which will reverse the order ($log_1/xy=-log_x y$).
We have
$$log_327=3,$$
$$log_54<1,$$
because $4<5^1$ and
$$1<log_25<3$$
because $2^1<5<2^3$.
The rest is yours.
Short answer:
$$log_54<1<log_25<3=log_327$$
and
$$log_1/54>log_1/25>log_1/327.$$
answered 4 hours ago
Yves Daoust
114k666209
For clarity, we take the inverses of the bases, which will reverse the order ($log_1/xy=-log_x y$).
We have
$$log_327=3,$$
$$log_54<1,$$
because $4<5^1$ and
$$1<log_25<3$$
because $2^1<5<2^3$.
The rest is yours.
Short answer:
$$log_54<1<log_25<3=log_327$$
and
$$log_1/54>log_1/25>log_1/327.$$
answered 4 hours ago
Yves Daoust
114k666209
edited 4 hours ago
answered 4 hours ago
Yves Daoust
114k666209
answered 4 hours ago
Yves Daoust
114k666209
answered 4 hours ago
Yves Daoust
114k666209
114k666209
add a comment |Â
add a comment |Â
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Quite correct, though you can make it simpler.
– Yves Daoust
4 hours ago