Find $log _2448$ if $log_1236=k$

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Find $log _2448$ if $log_1236=k$

My method:

We have $$fraclog 36log 12=k$$ $implies$

$$fraclog 12+log 3log 12=k$$ $implies$

$$fraclog32log 2+log 3=k-1$$ So

$$log 3=(k-1)t tag1$$

$$2log 2+log 3=t$$ $implies$

$$log 2=frac(2-k)t2 tag2$$

Now $$log _2448=fraclog 48log 24=frac4log 2+log 33log 2+log 3=frac2(2-k)+k-13left(frac2-k2right)+k-1=frac6-2k4-k$$

is there any other approach?

algebra-precalculus logarithms exponential-function ratio

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asked Jul 20 at 18:50

Umesh shankar

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up vote
4
down vote

favorite

1

Find $log _2448$ if $log_1236=k$

My method:

We have $$fraclog 36log 12=k$$ $implies$

$$fraclog 12+log 3log 12=k$$ $implies$

$$fraclog32log 2+log 3=k-1$$ So

$$log 3=(k-1)t tag1$$

$$2log 2+log 3=t$$ $implies$

$$log 2=frac(2-k)t2 tag2$$

Now $$log _2448=fraclog 48log 24=frac4log 2+log 33log 2+log 3=frac2(2-k)+k-13left(frac2-k2right)+k-1=frac6-2k4-k$$

is there any other approach?

algebra-precalculus logarithms exponential-function ratio

share|cite|improve this question

asked Jul 20 at 18:50

Umesh shankar

2,28611018

add a comment | 

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

Find $log _2448$ if $log_1236=k$

My method:

We have $$fraclog 36log 12=k$$ $implies$

$$fraclog 12+log 3log 12=k$$ $implies$

$$fraclog32log 2+log 3=k-1$$ So

$$log 3=(k-1)t tag1$$

$$2log 2+log 3=t$$ $implies$

$$log 2=frac(2-k)t2 tag2$$

Now $$log _2448=fraclog 48log 24=frac4log 2+log 33log 2+log 3=frac2(2-k)+k-13left(frac2-k2right)+k-1=frac6-2k4-k$$

is there any other approach?

algebra-precalculus logarithms exponential-function ratio

share|cite|improve this question

asked Jul 20 at 18:50

Umesh shankar

2,28611018

Find $log _2448$ if $log_1236=k$

My method:

We have $$fraclog 36log 12=k$$ $implies$

$$fraclog 12+log 3log 12=k$$ $implies$

$$fraclog32log 2+log 3=k-1$$ So

$$log 3=(k-1)t tag1$$

$$2log 2+log 3=t$$ $implies$

$$log 2=frac(2-k)t2 tag2$$

Now $$log _2448=fraclog 48log 24=frac4log 2+log 33log 2+log 3=frac2(2-k)+k-13left(frac2-k2right)+k-1=frac6-2k4-k$$

is there any other approach?

algebra-precalculus logarithms exponential-function ratio

share|cite|improve this question

asked Jul 20 at 18:50

Umesh shankar

2,28611018

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 18:50

Umesh shankar

2,28611018

asked Jul 20 at 18:50

Umesh shankar

2,28611018

asked Jul 20 at 18:50

Umesh shankar

2,28611018

2,28611018

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add a comment | 

4 Answers
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accepted

It seems easier to use $log_12$ rather than $log$.

$log_12(36)=log_12(3)+1=k$, and
$$2log_12(2)+log_12(3)=log_12(12)=1,$$ so $2log_12(2)=2-k$,
or $$log_12(2)=1-frack2$$

Thus
$$log_24(48)= fraclog_12(48)log_12(24)=frac2log_12(2)+1log_12(2)+1$$
$$=frac2-k+12-frack2=frac6-2k4-k.$$

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answered Jul 20 at 19:03

jgon

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With $z = log_2(3)$:

$$L = log_24(48) = fracln(48)ln(24) = fracln(2^4times 3)ln(2^3 times 3) = frac4ln(2) + ln(3)3ln(2) + ln(3) = frac4+z3 + z$$
$$K = log_12(36) = fracln(36)ln(12) = fracln(2^2times 3^2)ln(2^2 times 3) = frac2ln(2) + 2ln(3)2ln(2) + ln(3) = frac2+2z2 + z$$

Just solve for $z$ and plug:

$$z = frac2K-2K-2$$
$$L = frac2K - 6K-4$$

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answered Jul 20 at 19:17

DanielV

17.4k42651

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We have

$$log _2448=fraclog _1248log _1224=fraclog _1236+log _12frac43log _1236+log _12frac23=frack+log _12frac43k+log _12frac23$$

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edited Jul 20 at 19:49

answered Jul 20 at 19:42

gimusi

65.4k73584

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You can convert all to the smallest common base $2$:
$$log_1236=fraclog_236log_2 12=frac2+2log_2 32+log_2 3=k Rightarrow log_2 3=frac2k-22-k.$$
Hence:
$$log _2448=fraclog_2 48log_2 24=frac4+log_233+log_23=frac4+frac2k-22-k3+frac2k-22-k=frac6-2k4-k.$$

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answered Jul 21 at 13:04

farruhota

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

It seems easier to use $log_12$ rather than $log$.

$log_12(36)=log_12(3)+1=k$, and
$$2log_12(2)+log_12(3)=log_12(12)=1,$$ so $2log_12(2)=2-k$,
or $$log_12(2)=1-frack2$$

Thus
$$log_24(48)= fraclog_12(48)log_12(24)=frac2log_12(2)+1log_12(2)+1$$
$$=frac2-k+12-frack2=frac6-2k4-k.$$

share|cite|improve this answer

answered Jul 20 at 19:03

jgon

8,54611435

add a comment | 

up vote
4
down vote



accepted

It seems easier to use $log_12$ rather than $log$.

$log_12(36)=log_12(3)+1=k$, and
$$2log_12(2)+log_12(3)=log_12(12)=1,$$ so $2log_12(2)=2-k$,
or $$log_12(2)=1-frack2$$

Thus
$$log_24(48)= fraclog_12(48)log_12(24)=frac2log_12(2)+1log_12(2)+1$$
$$=frac2-k+12-frack2=frac6-2k4-k.$$

share|cite|improve this answer

answered Jul 20 at 19:03

jgon

8,54611435

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

It seems easier to use $log_12$ rather than $log$.

$log_12(36)=log_12(3)+1=k$, and
$$2log_12(2)+log_12(3)=log_12(12)=1,$$ so $2log_12(2)=2-k$,
or $$log_12(2)=1-frack2$$

Thus
$$log_24(48)= fraclog_12(48)log_12(24)=frac2log_12(2)+1log_12(2)+1$$
$$=frac2-k+12-frack2=frac6-2k4-k.$$

share|cite|improve this answer

answered Jul 20 at 19:03

jgon

8,54611435

It seems easier to use $log_12$ rather than $log$.

$log_12(36)=log_12(3)+1=k$, and
$$2log_12(2)+log_12(3)=log_12(12)=1,$$ so $2log_12(2)=2-k$,
or $$log_12(2)=1-frack2$$

Thus
$$log_24(48)= fraclog_12(48)log_12(24)=frac2log_12(2)+1log_12(2)+1$$
$$=frac2-k+12-frack2=frac6-2k4-k.$$

share|cite|improve this answer

answered Jul 20 at 19:03

jgon

8,54611435

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 19:03

jgon

8,54611435

answered Jul 20 at 19:03

jgon

8,54611435

answered Jul 20 at 19:03

jgon

8,54611435

8,54611435

add a comment | 

add a comment | 

up vote
0
down vote

With $z = log_2(3)$:

$$L = log_24(48) = fracln(48)ln(24) = fracln(2^4times 3)ln(2^3 times 3) = frac4ln(2) + ln(3)3ln(2) + ln(3) = frac4+z3 + z$$
$$K = log_12(36) = fracln(36)ln(12) = fracln(2^2times 3^2)ln(2^2 times 3) = frac2ln(2) + 2ln(3)2ln(2) + ln(3) = frac2+2z2 + z$$

Just solve for $z$ and plug:

$$z = frac2K-2K-2$$
$$L = frac2K - 6K-4$$

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answered Jul 20 at 19:17

DanielV

17.4k42651

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up vote
0
down vote

With $z = log_2(3)$:

$$L = log_24(48) = fracln(48)ln(24) = fracln(2^4times 3)ln(2^3 times 3) = frac4ln(2) + ln(3)3ln(2) + ln(3) = frac4+z3 + z$$
$$K = log_12(36) = fracln(36)ln(12) = fracln(2^2times 3^2)ln(2^2 times 3) = frac2ln(2) + 2ln(3)2ln(2) + ln(3) = frac2+2z2 + z$$

Just solve for $z$ and plug:

$$z = frac2K-2K-2$$
$$L = frac2K - 6K-4$$

share|cite|improve this answer

answered Jul 20 at 19:17

DanielV

17.4k42651

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up vote
0
down vote

up vote
0
down vote

With $z = log_2(3)$:

$$L = log_24(48) = fracln(48)ln(24) = fracln(2^4times 3)ln(2^3 times 3) = frac4ln(2) + ln(3)3ln(2) + ln(3) = frac4+z3 + z$$
$$K = log_12(36) = fracln(36)ln(12) = fracln(2^2times 3^2)ln(2^2 times 3) = frac2ln(2) + 2ln(3)2ln(2) + ln(3) = frac2+2z2 + z$$

Just solve for $z$ and plug:

$$z = frac2K-2K-2$$
$$L = frac2K - 6K-4$$

share|cite|improve this answer

answered Jul 20 at 19:17

DanielV

17.4k42651

With $z = log_2(3)$:

$$L = log_24(48) = fracln(48)ln(24) = fracln(2^4times 3)ln(2^3 times 3) = frac4ln(2) + ln(3)3ln(2) + ln(3) = frac4+z3 + z$$
$$K = log_12(36) = fracln(36)ln(12) = fracln(2^2times 3^2)ln(2^2 times 3) = frac2ln(2) + 2ln(3)2ln(2) + ln(3) = frac2+2z2 + z$$

Just solve for $z$ and plug:

$$z = frac2K-2K-2$$
$$L = frac2K - 6K-4$$

share|cite|improve this answer

answered Jul 20 at 19:17

DanielV

17.4k42651

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 19:17

DanielV

17.4k42651

answered Jul 20 at 19:17

DanielV

17.4k42651

answered Jul 20 at 19:17

DanielV

17.4k42651

17.4k42651

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add a comment | 

up vote
0
down vote

We have

$$log _2448=fraclog _1248log _1224=fraclog _1236+log _12frac43log _1236+log _12frac23=frack+log _12frac43k+log _12frac23$$

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edited Jul 20 at 19:49

answered Jul 20 at 19:42

gimusi

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up vote
0
down vote

We have

$$log _2448=fraclog _1248log _1224=fraclog _1236+log _12frac43log _1236+log _12frac23=frack+log _12frac43k+log _12frac23$$

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edited Jul 20 at 19:49

answered Jul 20 at 19:42

gimusi

65.4k73584

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up vote
0
down vote

up vote
0
down vote

We have

$$log _2448=fraclog _1248log _1224=fraclog _1236+log _12frac43log _1236+log _12frac23=frack+log _12frac43k+log _12frac23$$

share|cite|improve this answer

edited Jul 20 at 19:49

answered Jul 20 at 19:42

gimusi

65.4k73584

We have

$$log _2448=fraclog _1248log _1224=fraclog _1236+log _12frac43log _1236+log _12frac23=frack+log _12frac43k+log _12frac23$$

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edited Jul 20 at 19:49

answered Jul 20 at 19:42

gimusi

65.4k73584

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 19:49

edited Jul 20 at 19:49

edited Jul 20 at 19:49

answered Jul 20 at 19:42

gimusi

65.4k73584

answered Jul 20 at 19:42

gimusi

65.4k73584

answered Jul 20 at 19:42

gimusi

65.4k73584

65.4k73584

add a comment | 

add a comment | 

up vote
0
down vote

You can convert all to the smallest common base $2$:
$$log_1236=fraclog_236log_2 12=frac2+2log_2 32+log_2 3=k Rightarrow log_2 3=frac2k-22-k.$$
Hence:
$$log _2448=fraclog_2 48log_2 24=frac4+log_233+log_23=frac4+frac2k-22-k3+frac2k-22-k=frac6-2k4-k.$$

share|cite|improve this answer

answered Jul 21 at 13:04

farruhota

13.7k2632

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up vote
0
down vote

You can convert all to the smallest common base $2$:
$$log_1236=fraclog_236log_2 12=frac2+2log_2 32+log_2 3=k Rightarrow log_2 3=frac2k-22-k.$$
Hence:
$$log _2448=fraclog_2 48log_2 24=frac4+log_233+log_23=frac4+frac2k-22-k3+frac2k-22-k=frac6-2k4-k.$$

share|cite|improve this answer

answered Jul 21 at 13:04

farruhota

13.7k2632

add a comment | 

up vote
0
down vote

up vote
0
down vote

You can convert all to the smallest common base $2$:
$$log_1236=fraclog_236log_2 12=frac2+2log_2 32+log_2 3=k Rightarrow log_2 3=frac2k-22-k.$$
Hence:
$$log _2448=fraclog_2 48log_2 24=frac4+log_233+log_23=frac4+frac2k-22-k3+frac2k-22-k=frac6-2k4-k.$$

share|cite|improve this answer

answered Jul 21 at 13:04

farruhota

13.7k2632

You can convert all to the smallest common base $2$:
$$log_1236=fraclog_236log_2 12=frac2+2log_2 32+log_2 3=k Rightarrow log_2 3=frac2k-22-k.$$
Hence:
$$log _2448=fraclog_2 48log_2 24=frac4+log_233+log_23=frac4+frac2k-22-k3+frac2k-22-k=frac6-2k4-k.$$

share|cite|improve this answer

answered Jul 21 at 13:04

farruhota

13.7k2632

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share|cite|improve this answer

answered Jul 21 at 13:04

farruhota

13.7k2632

answered Jul 21 at 13:04

farruhota

13.7k2632

answered Jul 21 at 13:04

farruhota

13.7k2632

13.7k2632

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add a comment | 

 

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