If $2^a=3^b$ find $frac{a}{b}$ [closed]

-1

I tried many different things but still couldn't solve it. Could you please give me a clue?

asked Dec 24 '18 at 18:00

Sebastian Akbari

closed as off-topic by Abcd, mrtaurho, amWhy, Saad, Jyrki Lahtonen Dec 28 '18 at 10:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Abcd, mrtaurho, amWhy, Saad, Jyrki Lahtonen

If this question can be reworded to fit the rules in the help center, please edit the question.

5

$begingroup$
logarithms?${}$
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 18:01

$begingroup$
Use logarithm..
$endgroup$
– Love Invariants
Dec 24 '18 at 18:01

$begingroup$
Hint: take $,ell = {rm log}_2,$ of both sides to get $, a = b,ell(3) $
$endgroup$
– Bill Dubuque
Dec 24 '18 at 19:50

1

$begingroup$
See also Prove that $log_2 3$ is irrational
$endgroup$
– Bill Dubuque
Dec 24 '18 at 20:08

add a comment |

-1

I tried many different things but still couldn't solve it. Could you please give me a clue?

asked Dec 24 '18 at 18:00

Sebastian Akbari

closed as off-topic by Abcd, mrtaurho, amWhy, Saad, Jyrki Lahtonen Dec 28 '18 at 10:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Abcd, mrtaurho, amWhy, Saad, Jyrki Lahtonen

If this question can be reworded to fit the rules in the help center, please edit the question.

5

$begingroup$
logarithms?${}$
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 18:01

$begingroup$
Use logarithm..
$endgroup$
– Love Invariants
Dec 24 '18 at 18:01

$begingroup$
Hint: take $,ell = {rm log}_2,$ of both sides to get $, a = b,ell(3) $
$endgroup$
– Bill Dubuque
Dec 24 '18 at 19:50

1

$begingroup$
See also Prove that $log_2 3$ is irrational
$endgroup$
– Bill Dubuque
Dec 24 '18 at 20:08

add a comment |

-1

I tried many different things but still couldn't solve it. Could you please give me a clue?

asked Dec 24 '18 at 18:00

Sebastian Akbari

I tried many different things but still couldn't solve it. Could you please give me a clue?

algebra-precalculus logarithms exponentiation

asked Dec 24 '18 at 18:00

Sebastian Akbari

asked Dec 24 '18 at 18:00

Sebastian Akbari

asked Dec 24 '18 at 18:00

Sebastian Akbari

asked Dec 24 '18 at 18:00

Sebastian Akbari

asked Dec 24 '18 at 18:00

Sebastian Akbari

closed as off-topic by Abcd, mrtaurho, amWhy, Saad, Jyrki Lahtonen Dec 28 '18 at 10:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Abcd, mrtaurho, amWhy, Saad, Jyrki Lahtonen

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Abcd, mrtaurho, amWhy, Saad, Jyrki Lahtonen Dec 28 '18 at 10:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Abcd, mrtaurho, amWhy, Saad, Jyrki Lahtonen

If this question can be reworded to fit the rules in the help center, please edit the question.

5

$begingroup$
logarithms?${}$
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 18:01

$begingroup$
Use logarithm..
$endgroup$
– Love Invariants
Dec 24 '18 at 18:01

$begingroup$
Hint: take $,ell = {rm log}_2,$ of both sides to get $, a = b,ell(3) $
$endgroup$
– Bill Dubuque
Dec 24 '18 at 19:50

1

$begingroup$
See also Prove that $log_2 3$ is irrational
$endgroup$
– Bill Dubuque
Dec 24 '18 at 20:08

add a comment |

5

$begingroup$
logarithms?${}$
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 18:01

$begingroup$
Use logarithm..
$endgroup$
– Love Invariants
Dec 24 '18 at 18:01

$begingroup$
Hint: take $,ell = {rm log}_2,$ of both sides to get $, a = b,ell(3) $
$endgroup$
– Bill Dubuque
Dec 24 '18 at 19:50

1

$begingroup$
See also Prove that $log_2 3$ is irrational
$endgroup$
– Bill Dubuque
Dec 24 '18 at 20:08

logarithms?${}$

– Lord Shark the Unknown
Dec 24 '18 at 18:01

Use logarithm..

– Love Invariants
Dec 24 '18 at 18:01

Hint: take $,ell = {rm log}_2,$ of both sides to get $, a = b,ell(3) $

– Bill Dubuque
Dec 24 '18 at 19:50

See also Prove that $log_2 3$ is irrational

– Bill Dubuque
Dec 24 '18 at 20:08

add a comment |

2 Answers
2

active

oldest

votes

Try taking the natural log of both sides. Remember that $ln 2^a=aln 2$ and $ln 3^b=bln 3$.

$$aln 2=bln 3$$

Now, you can use division to solve for $frac a b$.

answered Dec 24 '18 at 18:02

Noble Mushtak

15.3k1835

add a comment |

$2^a = 3^b; tag 1$

$ln (2^a) = ln (3^b); tag 2$

$a ln 2 = b ln 3; tag 3$

$dfrac{a}{b} = dfrac{ln 3}{ln 2}. tag 4$

answered Dec 24 '18 at 18:04

Robert Lewis

46k23066

add a comment |

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Try taking the natural log of both sides. Remember that $ln 2^a=aln 2$ and $ln 3^b=bln 3$.

$$aln 2=bln 3$$

Now, you can use division to solve for $frac a b$.

answered Dec 24 '18 at 18:02

Noble Mushtak

15.3k1835

add a comment |

Try taking the natural log of both sides. Remember that $ln 2^a=aln 2$ and $ln 3^b=bln 3$.

$$aln 2=bln 3$$

Now, you can use division to solve for $frac a b$.

answered Dec 24 '18 at 18:02

Noble Mushtak

15.3k1835

add a comment |

Try taking the natural log of both sides. Remember that $ln 2^a=aln 2$ and $ln 3^b=bln 3$.

$$aln 2=bln 3$$

Now, you can use division to solve for $frac a b$.

answered Dec 24 '18 at 18:02

Noble Mushtak

15.3k1835

Try taking the natural log of both sides. Remember that $ln 2^a=aln 2$ and $ln 3^b=bln 3$.

$$aln 2=bln 3$$

Now, you can use division to solve for $frac a b$.

answered Dec 24 '18 at 18:02

Noble Mushtak

15.3k1835

answered Dec 24 '18 at 18:02

Noble Mushtak

15.3k1835

answered Dec 24 '18 at 18:02

Noble Mushtak

15.3k1835

answered Dec 24 '18 at 18:02

Noble Mushtak

15.3k1835

add a comment |

$2^a = 3^b; tag 1$

$ln (2^a) = ln (3^b); tag 2$

$a ln 2 = b ln 3; tag 3$

$dfrac{a}{b} = dfrac{ln 3}{ln 2}. tag 4$

answered Dec 24 '18 at 18:04

Robert Lewis

46k23066

add a comment |

$2^a = 3^b; tag 1$

$ln (2^a) = ln (3^b); tag 2$

$a ln 2 = b ln 3; tag 3$

$dfrac{a}{b} = dfrac{ln 3}{ln 2}. tag 4$

answered Dec 24 '18 at 18:04

Robert Lewis

46k23066

add a comment |

$2^a = 3^b; tag 1$

$ln (2^a) = ln (3^b); tag 2$

$a ln 2 = b ln 3; tag 3$

$dfrac{a}{b} = dfrac{ln 3}{ln 2}. tag 4$

answered Dec 24 '18 at 18:04

Robert Lewis

46k23066

$2^a = 3^b; tag 1$

$ln (2^a) = ln (3^b); tag 2$

$a ln 2 = b ln 3; tag 3$

$dfrac{a}{b} = dfrac{ln 3}{ln 2}. tag 4$

answered Dec 24 '18 at 18:04

Robert Lewis

46k23066

answered Dec 24 '18 at 18:04

Robert Lewis

46k23066

answered Dec 24 '18 at 18:04

Robert Lewis

46k23066

answered Dec 24 '18 at 18:04

Robert Lewis

46k23066

add a comment |

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