Logarithms help

[itsDANNY]

Hi friends,

Given the formula to calculate the future value of a current amount, and rearranging the formula to calculate t, the number of periods, such that:

FV = PV(1+i)^t
FV/PV = (1+i)^t
Taking loge of both sides gives: loge(FV/PV) = loge(1+i)^t
Hence by one of the logarithmic laws: loge(FV/PV) = t*loge(1+i)
Thus: t = loge(FV/PV)/loge(1+i)

Why is it that we cannot cancel the two log of base e here and simplify down to t = (FV/PV) / (1+i)?

Both gives different answers and I'm just wondering why you cannot cancel or simplify out the loge?

Thanks.

 

[stapel]

FV = PV(1+i)^t
FV/PV = (1+i)^t
Taking loge of both sides gives: loge(FV/PV) = loge(1+i)^t
Hence by one of the logarithmic laws: loge(FV/PV) = t*loge(1+i)
Thus: t = loge(FV/PV)/loge(1+i)

Why is it that we cannot cancel the two log of base e here and simplify down to t = (FV/PV) / (1+i)?

Um... Because "log" is not a variable that can be "cancelled"; it's function which must be evaluated.

Did they not teach you about logarithms? :shock:

 

[HallsofIvy]

In general, if f is a function then \(\displaystyle \frac{f(x)}{f(y)}\) is NOT equal to \(\displaystyle \frac{x}{y}\). For example, in the very simple case of f(x)= 2x+ 5, f(1)= 2(1)+ 5= 7 and f(3)= 2(3)+ 5= 11 so \(\displaystyle \frac{f(2)}{f(3)}= \frac{7}{11}\) which is NOT equal to \(\displaystyle \frac{1}{3}\).

 

[itsDANNY]

Um... Because "log" is not a variable that can be "cancelled"; it's function which must be evaluated.

Did they not teach you about logarithms? :shock:

Great thanks that makes sense. And you're right, unfortunately not.

In general, if f is a function then \(\displaystyle \frac{f(x)}{f(y)}\) is NOT equal to \(\displaystyle \frac{x}{y}\). For example, in the very simple case of f(x)= 2x+ 5, f(1)= 2(1)+ 5= 7 and f(3)= 2(3)+ 5= 11 so \(\displaystyle \frac{f(2)}{f(3)}= \frac{7}{11}\) which is NOT equal to \(\displaystyle \frac{1}{3}\).

Sorry, but just to clarify, did you mean to say, where f(x)= 2x+ 5, \(\displaystyle \frac{f(1)}{f(3)}= \frac{7}{11}\) rather than \(\displaystyle \frac{f(2)}{f(3)}= \frac{7}{11}\) to illustrate your point?

Any ways, thank you for the prompt responses bodyevery.

Have a great day y'all.

 

[Dale10101]

Cancel that idea

Hi friends,

Given the formula to calculate the future value of a current amount, and rearranging the formula to calculate t, the number of periods, such that:

FV = PV(1+i)^t
FV/PV = (1+i)^t
Taking loge of both sides gives: loge(FV/PV) = loge(1+i)^t
Hence by one of the logarithmic laws: loge(FV/PV) = t*loge(1+i)
Thus: t = loge(FV/PV)/loge(1+i)

Why is it that we cannot cancel the two log of base e here and simplify down to t = (FV/PV) / (1+i)?

Both gives different answers and I'm just wondering why you cannot cancel or simplify out the loge?

Thanks.

We cancel to reduce a ratio, i.e

c = 100/10 = 10/1,

s = 2(100)/(2(10) = 10/1

The application of a log function to a numerator and denominator does not step each up or down by the same ratio,

log (100) / log(10) = 2/1 .

log( 2(100) ) / log( 2(10) ) = 2.30/1

One can think of a function's stimulus input as the length of the number line from zero to x along the x-axis, and its output response as the length of the number line from zero to y along the orthogonal y axis.

If you compare the input and output lengths of a function for different values of x and find a common ratio of lengths then you have something in common to cancel in a ratio between the two, if not, as with log functions, there in nothing in common to cancel.

Finally, if y = log(x), then the output per input comparison between y and x with respect to a ratio of the two is: log(x)/x. Graphing this function shows that there is no common ratio to cancel.


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[stapel]

Did they not teach you about logarithms?

...unfortunately not.

Ouch! You can study at least some of the basics of logarithms online, such as in this listing.

 

[lookagain]

The application of a log function to a numerator and denominator does not step each up or down
by the same ratio,

log (100) / log(10) = 2/1 .

log( 2(100) ) / log( 2(10) ) = 2.30/1

This is a correction:

log(2(100))/log(2(10)) \(\displaystyle \ \approx \ \) 2.30/1.30

 

[Dale10101]

Exactly so.

This is a correction:

log(2(100))/log(2(10)) \(\displaystyle \ \approx \ \) 2.30/1.30

Thanks.