How do I find the value of N, 2 = (1.07)^N

[georgemathew]

find N

2 = {1.07}^{N}

I have the answer in my booklet as 10.24. However, kindly help me with the steps to derive the answer

 

[Deleted member 4993]

find N

2 = {1.07}^{N}

I have the answer in my booklet as 10.24. However, kindly help me with the steps to derive the answer

Use the law of logarithm:

log(a^b) = b * log(a)

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.

 

[pka]

If [imath]2=(1.07)^N[/imath] Then [imath]\log(2)=N\log(1.07)[/imath] or [imath]N=\dfrac{\log(2)}{\log(1.07)}[/imath]
SEE HERE.


[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

 

[Harry_the_cat]

If you haven't learnt about logarithms, then you would probably be expected to solve it by trial and error. Or graphically.

 

[georgemathew]

If [imath]2=(1.07)^N[/imath] Then [imath]\log(2)=N\log(1.07)[/imath] or [imath]N=\dfrac{\log(2)}{\log(1.07)}[/imath]
SEE HERE.


[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

Thanks! I found the steps and the solution

 

[Steven G]

If you haven't learnt about logarithms, then you would probably be expected to solve it by trial and error. Or graphically.

I would hope that a student studying differential equations has encountered logarithms at some point.

 

[Harry_the_cat]

I would hope that a student studying differential equations has encountered logarithms at some point.

You would hope so.