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Solve in terms of $ln(5)$ and $ln(3)$: $15^4y=3^4y+8$

Finding the inverse of a log problemExpressing logarithms as ratios of natural logarithmsA train traveled at a constant rate of $f$ feet per second. How many feet did it travel in $x$ minutes?What's wrong with solving absolute value equations in this way?Solve the algebra equation- unsure about order of operations, how to go about solving, solve for xSolving $sin fractheta2 + cos fractheta2 = sqrt2$Intersection of a line and a curve. Why is my reasoning incorrect?Why doesn't $77=1$? (I'll explain in question)Expanding log problemFor a certain value of $p$, point $T_p$ lies on the line $y=2x$

1

$begingroup$

Here is my attempt, and I don't understand why it doesn't work (I apologise for my clunky MathJax):

$(4y)ln15=(4y+8)ln3$

$(4y)(ln3+ln5)=(4y+8)ln3$

$4(y(ln3+ln5)=4(y+2ln3)$

$y(ln3+ln5)=(y+2ln3)$

$y(ln3+ln5)-y=(2ln3)$

$y(ln3+ln5-1)=(2ln3)$

$y=frac2ln3ln3+ln5-1$

The correct answer, however, is $frac2ln3ln5$. Could someone point to my mistake, as I can't recognise it myself? Thank you.

algebra-precalculus logarithms

share|cite|improve this question

edited Mar 21 at 13:34

Adam Grey

asked Mar 21 at 13:22

Adam GreyAdam Grey

484

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If the original problem is indeed $(ln 15)^4y=(ln 3)^4y+8$, the next line should then be $$4yln(ln3+ln5)=(4y+8)ln(ln 3).$$ The only way the second line is $4yln15=(4y+8)ln3$ is if the original problem was instead $15^4y=3^4y+8$. You should check if your title is correct or not.
  $endgroup$
  – arctic tern
  Mar 21 at 13:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @arctictern You are correct, the title is wrong. Thank you and I apologise for the confusion.
  $endgroup$
  – Adam Grey
  Mar 21 at 13:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Typesetting hint: put a backslash before ln to make it display in roman type (ln x gives $ln x$, while ln x gives $ln x$). The same goes for sin, cos, etc.
  $endgroup$
  – Théophile
  Mar 21 at 13:40
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Here is my attempt, and I don't understand why it doesn't work (I apologise for my clunky MathJax):

$(4y)ln15=(4y+8)ln3$

$(4y)(ln3+ln5)=(4y+8)ln3$

$4(y(ln3+ln5)=4(y+2ln3)$

$y(ln3+ln5)=(y+2ln3)$

$y(ln3+ln5)-y=(2ln3)$

$y(ln3+ln5-1)=(2ln3)$

$y=frac2ln3ln3+ln5-1$

The correct answer, however, is $frac2ln3ln5$. Could someone point to my mistake, as I can't recognise it myself? Thank you.

algebra-precalculus logarithms

share|cite|improve this question

edited Mar 21 at 13:34

Adam Grey

asked Mar 21 at 13:22

Adam GreyAdam Grey

484

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If the original problem is indeed $(ln 15)^4y=(ln 3)^4y+8$, the next line should then be $$4yln(ln3+ln5)=(4y+8)ln(ln 3).$$ The only way the second line is $4yln15=(4y+8)ln3$ is if the original problem was instead $15^4y=3^4y+8$. You should check if your title is correct or not.
  $endgroup$
  – arctic tern
  Mar 21 at 13:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @arctictern You are correct, the title is wrong. Thank you and I apologise for the confusion.
  $endgroup$
  – Adam Grey
  Mar 21 at 13:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Typesetting hint: put a backslash before ln to make it display in roman type (ln x gives $ln x$, while ln x gives $ln x$). The same goes for sin, cos, etc.
  $endgroup$
  – Théophile
  Mar 21 at 13:40
  
  

  
  
  
  

add a comment | 

$begingroup$

Here is my attempt, and I don't understand why it doesn't work (I apologise for my clunky MathJax):

$(4y)ln15=(4y+8)ln3$

$(4y)(ln3+ln5)=(4y+8)ln3$

$4(y(ln3+ln5)=4(y+2ln3)$

$y(ln3+ln5)=(y+2ln3)$

$y(ln3+ln5)-y=(2ln3)$

$y(ln3+ln5-1)=(2ln3)$

$y=frac2ln3ln3+ln5-1$

The correct answer, however, is $frac2ln3ln5$. Could someone point to my mistake, as I can't recognise it myself? Thank you.

algebra-precalculus logarithms

share|cite|improve this question

edited Mar 21 at 13:34

Adam Grey

asked Mar 21 at 13:22

Adam GreyAdam Grey

484

$endgroup$

share|cite|improve this question

edited Mar 21 at 13:34

Adam Grey

asked Mar 21 at 13:22

Adam GreyAdam Grey

484

share|cite|improve this question

edited Mar 21 at 13:34

Adam Grey

asked Mar 21 at 13:22

Adam GreyAdam Grey

484

asked Mar 21 at 13:22

Adam GreyAdam Grey

484

asked Mar 21 at 13:22

Adam GreyAdam Grey

484

3 Answers
3

active

oldest

votes

1

$begingroup$

Here is the mistake:

$$4y(ln3+ln5)=4(y+2colorred)ln3$$

$$y(ln 3 + ln 5) = (y+2) ln 3$$

$$yln 5 = 2ln 3$$

share|cite|improve this answer

answered Mar 21 at 13:26

Siong Thye GohSiong Thye Goh

103k1468120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you interpreting the original problem as $15^4y=3^4y+8$? If so, you should be clear and explicit when you believe an OP has incorrectly posed a problem and how you're reinterpreting it.
  $endgroup$
  – arctic tern
  Mar 21 at 13:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am intepreting it as $15^4y=3^4y+8$.
  $endgroup$
  – Siong Thye Goh
  Mar 21 at 13:30
  

  
  
  
  

add a comment | 

2

$begingroup$

Mistake is in step $2$ and $3$ $$(4y+8)ln 3ne 4(y+2ln3)$$

share|cite|improve this answer

answered Mar 21 at 13:33

Paras KhoslaParas Khosla

2,792423

$endgroup$

add a comment | 

1

$begingroup$

i think it must be $$4y(ln(3)+ln(5))=(4y+8)ln(3)$$

share|cite|improve this answer

edited Mar 21 at 13:37

answered Mar 21 at 13:29

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

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1

$begingroup$

Here is the mistake:

$$4y(ln3+ln5)=4(y+2colorred)ln3$$

$$y(ln 3 + ln 5) = (y+2) ln 3$$

$$yln 5 = 2ln 3$$

share|cite|improve this answer

answered Mar 21 at 13:26

Siong Thye GohSiong Thye Goh

103k1468120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you interpreting the original problem as $15^4y=3^4y+8$? If so, you should be clear and explicit when you believe an OP has incorrectly posed a problem and how you're reinterpreting it.
  $endgroup$
  – arctic tern
  Mar 21 at 13:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am intepreting it as $15^4y=3^4y+8$.
  $endgroup$
  – Siong Thye Goh
  Mar 21 at 13:30
  

  
  
  
  

add a comment | 

1

$begingroup$

Here is the mistake:

$$4y(ln3+ln5)=4(y+2colorred)ln3$$

$$y(ln 3 + ln 5) = (y+2) ln 3$$

$$yln 5 = 2ln 3$$

share|cite|improve this answer

answered Mar 21 at 13:26

Siong Thye GohSiong Thye Goh

103k1468120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you interpreting the original problem as $15^4y=3^4y+8$? If so, you should be clear and explicit when you believe an OP has incorrectly posed a problem and how you're reinterpreting it.
  $endgroup$
  – arctic tern
  Mar 21 at 13:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am intepreting it as $15^4y=3^4y+8$.
  $endgroup$
  – Siong Thye Goh
  Mar 21 at 13:30
  

  
  
  
  

add a comment | 

$begingroup$

Here is the mistake:

$$4y(ln3+ln5)=4(y+2colorred)ln3$$

$$y(ln 3 + ln 5) = (y+2) ln 3$$

$$yln 5 = 2ln 3$$

share|cite|improve this answer

answered Mar 21 at 13:26

Siong Thye GohSiong Thye Goh

103k1468120

$endgroup$

share|cite|improve this answer

answered Mar 21 at 13:26

Siong Thye GohSiong Thye Goh

103k1468120

answered Mar 21 at 13:26

Siong Thye GohSiong Thye Goh

103k1468120

answered Mar 21 at 13:26

Siong Thye GohSiong Thye Goh

103k1468120

2

$begingroup$

Mistake is in step $2$ and $3$ $$(4y+8)ln 3ne 4(y+2ln3)$$

share|cite|improve this answer

answered Mar 21 at 13:33

Paras KhoslaParas Khosla

2,792423

$endgroup$

add a comment | 

2

$begingroup$

Mistake is in step $2$ and $3$ $$(4y+8)ln 3ne 4(y+2ln3)$$

share|cite|improve this answer

answered Mar 21 at 13:33

Paras KhoslaParas Khosla

2,792423

$endgroup$

add a comment | 

$begingroup$

Mistake is in step $2$ and $3$ $$(4y+8)ln 3ne 4(y+2ln3)$$

share|cite|improve this answer

answered Mar 21 at 13:33

Paras KhoslaParas Khosla

2,792423

$endgroup$

share|cite|improve this answer

answered Mar 21 at 13:33

Paras KhoslaParas Khosla

2,792423

answered Mar 21 at 13:33

Paras KhoslaParas Khosla

2,792423

answered Mar 21 at 13:33

Paras KhoslaParas Khosla

2,792423

1

$begingroup$

i think it must be $$4y(ln(3)+ln(5))=(4y+8)ln(3)$$

share|cite|improve this answer

edited Mar 21 at 13:37

answered Mar 21 at 13:29

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

1

$begingroup$

i think it must be $$4y(ln(3)+ln(5))=(4y+8)ln(3)$$

share|cite|improve this answer

edited Mar 21 at 13:37

answered Mar 21 at 13:29

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

add a comment | 

$begingroup$

i think it must be $$4y(ln(3)+ln(5))=(4y+8)ln(3)$$

share|cite|improve this answer

edited Mar 21 at 13:37

answered Mar 21 at 13:29

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

$endgroup$

share|cite|improve this answer

edited Mar 21 at 13:37

answered Mar 21 at 13:29

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

answered Mar 21 at 13:29

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867

answered Mar 21 at 13:29

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.4k42867