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Solving for $n$ in $frac1.1^n - 1(0.1)1.1^n =4.6$

0

$begingroup$

I want to solve for $n$ in this equation:
$$PWF = frac(1+i)^n-1i(1+i)^n = frac1.1^n - 10.1 cdot 1.1^n =4.6$$

I know that log must be taken of both the right as well as the left sides of the equation but I am confused as to how.

algebra-precalculus logarithms

share|cite|improve this question

edited Apr 1 at 6:50

Eevee Trainer

10.5k31842

asked Apr 1 at 6:27

7bottlesofrum7bottlesofrum

6

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Cross multiply and find the value of $(1.1)^n$ first.
  $endgroup$
  – Kavi Rama Murthy
  Apr 1 at 6:31
  

  
  
  
  

add a comment | 

0

$begingroup$

I want to solve for $n$ in this equation:
$$PWF = frac(1+i)^n-1i(1+i)^n = frac1.1^n - 10.1 cdot 1.1^n =4.6$$

I know that log must be taken of both the right as well as the left sides of the equation but I am confused as to how.

algebra-precalculus logarithms

share|cite|improve this question

edited Apr 1 at 6:50

Eevee Trainer

10.5k31842

asked Apr 1 at 6:27

7bottlesofrum7bottlesofrum

6

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Cross multiply and find the value of $(1.1)^n$ first.
  $endgroup$
  – Kavi Rama Murthy
  Apr 1 at 6:31
  

  
  
  
  

add a comment | 

$begingroup$

I want to solve for $n$ in this equation:
$$PWF = frac(1+i)^n-1i(1+i)^n = frac1.1^n - 10.1 cdot 1.1^n =4.6$$

I know that log must be taken of both the right as well as the left sides of the equation but I am confused as to how.

algebra-precalculus logarithms

share|cite|improve this question

edited Apr 1 at 6:50

Eevee Trainer

10.5k31842

asked Apr 1 at 6:27

7bottlesofrum7bottlesofrum

6

$endgroup$

share|cite|improve this question

edited Apr 1 at 6:50

Eevee Trainer

10.5k31842

asked Apr 1 at 6:27

7bottlesofrum7bottlesofrum

6

share|cite|improve this question

edited Apr 1 at 6:50

Eevee Trainer

10.5k31842

asked Apr 1 at 6:27

7bottlesofrum7bottlesofrum

6

edited Apr 1 at 6:50

Eevee Trainer

10.5k31842

edited Apr 1 at 6:50

Eevee Trainer

10.5k31842

asked Apr 1 at 6:27

7bottlesofrum7bottlesofrum

6

asked Apr 1 at 6:27

7bottlesofrum7bottlesofrum

6

2 Answers
2

active

oldest

votes

0

$begingroup$

So, what we have is

$$frac1.1^n - 10.1 cdot 1.1^n =4.6$$

We multiply both sides by $0.1$ and split up the fraction:

$$frac1.1^n - 11.1^n = 1- frac11.1^n = 0.46$$

Solving for the fraction, we get

$$0.54 = frac11.1^n$$

Taking the reciprocal,

$$1.1^n = frac10.54 = frac5027$$

We take the logarithm, base $1.1$, of both sides.

$$log_1.1(1.1^n) = n = log_1.1 left( frac5027 right)$$

From there you pretty much have to plug it into a calculator to get a value. WolframAlpha gives about $6.46506$.

share|cite|improve this answer

answered Apr 1 at 6:46

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

add a comment | 

0

$begingroup$

Guide:

Try not to think of you have to apply logarithm from the start. Focus on the simplication and only take logarithm when the form is right.

Let $1.1^n=x$, now multiply both sides by the denominator on the right and we obtain a linear equation. Solve for $x$.

After solving for $x$, the problem becomes $1.1^n=x$ and we have $n=fracln xln 1.1$.

share|cite|improve this answer

edited Apr 1 at 8:45

answered Apr 1 at 7:30

Siong Thye GohSiong Thye Goh

104k1468120

$endgroup$

add a comment | 

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0

$begingroup$

So, what we have is

$$frac1.1^n - 10.1 cdot 1.1^n =4.6$$

We multiply both sides by $0.1$ and split up the fraction:

$$frac1.1^n - 11.1^n = 1- frac11.1^n = 0.46$$

Solving for the fraction, we get

$$0.54 = frac11.1^n$$

Taking the reciprocal,

$$1.1^n = frac10.54 = frac5027$$

We take the logarithm, base $1.1$, of both sides.

$$log_1.1(1.1^n) = n = log_1.1 left( frac5027 right)$$

From there you pretty much have to plug it into a calculator to get a value. WolframAlpha gives about $6.46506$.

share|cite|improve this answer

answered Apr 1 at 6:46

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

add a comment | 

0

$begingroup$

So, what we have is

$$frac1.1^n - 10.1 cdot 1.1^n =4.6$$

We multiply both sides by $0.1$ and split up the fraction:

$$frac1.1^n - 11.1^n = 1- frac11.1^n = 0.46$$

Solving for the fraction, we get

$$0.54 = frac11.1^n$$

Taking the reciprocal,

$$1.1^n = frac10.54 = frac5027$$

We take the logarithm, base $1.1$, of both sides.

$$log_1.1(1.1^n) = n = log_1.1 left( frac5027 right)$$

From there you pretty much have to plug it into a calculator to get a value. WolframAlpha gives about $6.46506$.

share|cite|improve this answer

answered Apr 1 at 6:46

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

add a comment | 

$begingroup$

So, what we have is

$$frac1.1^n - 10.1 cdot 1.1^n =4.6$$

We multiply both sides by $0.1$ and split up the fraction:

$$frac1.1^n - 11.1^n = 1- frac11.1^n = 0.46$$

Solving for the fraction, we get

$$0.54 = frac11.1^n$$

Taking the reciprocal,

$$1.1^n = frac10.54 = frac5027$$

We take the logarithm, base $1.1$, of both sides.

$$log_1.1(1.1^n) = n = log_1.1 left( frac5027 right)$$

From there you pretty much have to plug it into a calculator to get a value. WolframAlpha gives about $6.46506$.

share|cite|improve this answer

answered Apr 1 at 6:46

Eevee TrainerEevee Trainer

10.5k31842

$endgroup$

share|cite|improve this answer

answered Apr 1 at 6:46

Eevee TrainerEevee Trainer

10.5k31842

answered Apr 1 at 6:46

Eevee TrainerEevee Trainer

10.5k31842

answered Apr 1 at 6:46

Eevee TrainerEevee Trainer

10.5k31842

0

$begingroup$

Guide:

Try not to think of you have to apply logarithm from the start. Focus on the simplication and only take logarithm when the form is right.

Let $1.1^n=x$, now multiply both sides by the denominator on the right and we obtain a linear equation. Solve for $x$.

After solving for $x$, the problem becomes $1.1^n=x$ and we have $n=fracln xln 1.1$.

share|cite|improve this answer

edited Apr 1 at 8:45

answered Apr 1 at 7:30

Siong Thye GohSiong Thye Goh

104k1468120

$endgroup$

add a comment | 

0

$begingroup$

Guide:

Try not to think of you have to apply logarithm from the start. Focus on the simplication and only take logarithm when the form is right.

Let $1.1^n=x$, now multiply both sides by the denominator on the right and we obtain a linear equation. Solve for $x$.

After solving for $x$, the problem becomes $1.1^n=x$ and we have $n=fracln xln 1.1$.

share|cite|improve this answer

edited Apr 1 at 8:45

answered Apr 1 at 7:30

Siong Thye GohSiong Thye Goh

104k1468120

$endgroup$

add a comment | 

$begingroup$

Guide:

Try not to think of you have to apply logarithm from the start. Focus on the simplication and only take logarithm when the form is right.

Let $1.1^n=x$, now multiply both sides by the denominator on the right and we obtain a linear equation. Solve for $x$.

After solving for $x$, the problem becomes $1.1^n=x$ and we have $n=fracln xln 1.1$.

share|cite|improve this answer

edited Apr 1 at 8:45

answered Apr 1 at 7:30

Siong Thye GohSiong Thye Goh

104k1468120

$endgroup$

share|cite|improve this answer

edited Apr 1 at 8:45

answered Apr 1 at 7:30

Siong Thye GohSiong Thye Goh

104k1468120

answered Apr 1 at 7:30

Siong Thye GohSiong Thye Goh

104k1468120

answered Apr 1 at 7:30

Siong Thye GohSiong Thye Goh

104k1468120