Natural logarithm properties

[dollyayesha2345]

Question: Use the properties of natural logarithms to simplify the expression.
$$\ln e^{1/3}$$My answer: Inverse property should be used because
$$\ln e^{1/3}=\ln (1/e^3)$$
Is this correct? if not then which property should be used?

 

[pka]

Question: Use the properties of natural logarithms to simplify the expression.
$$\ln e^{1/3}$$

The correct anawer is $\large \dfrac{1}{3}$ Please explain WHY?

 

[JeffM]

WHOA

$$ln(e^{1/3}) \ne ln \left ( \dfrac{1}{e^3} \right ).$$

 

[dollyayesha2345]

WHOA

$$ln(e^{1/3}) \ne ln \left ( \dfrac{1}{e^3} \right ).$$

okay yes, absolutely my bad. The example said$$\ln (1/e^2) = \ln e^{-2}=-2$$ I got distracted by the fraction.
Please do help me though!

 

[BigBeachBanana]

You can review log and exponents here from the home page

Derivatives of Logs and Exponentials - Free Math Help

 

[dollyayesha2345]

You can review log and exponents here from the home page

Derivatives of Logs and Exponentials - Free Math Help

This is related to the differentiation of logs and I need to identify the "natural logarithm property".

 

[BigBeachBanana]

This is related to the differentiation of logs and I need to identify the "natural logarithm property".

Have you open the link? There’s review before talking about derivatives.

 

[Deleted member 4993]

Question: Use the properties of natural logarithms to simplify the expression.
$$\ln e^{1/3}$$My answer: Inverse property should be used because
$$\ln e^{1/3}=\ln (1/e^3)$$
Is this correct? if not then which property should be used?

Do you know the following property of logarithm:

$\displaystyle \log_a(b^m) = m * \log_a(b)$

 

[dollyayesha2345]

Do you know the following property of logarithm:

$\displaystyle \log_a(b^m) = m * \log_a(b)$

umm.. if I'm not wrong this is called "Power property"

 

[BigBeachBanana]

u

umm.. if I'm not wrong this is called "Power property"

Yes, Now identify m, a, and b. Then apply the "Power Rule".

 

[dollyayesha2345]

Yes, Now identify m, a, and b. Then apply the "Power Rule".

Expanding $$\ln e^{1/3}$$$$\frac{1}{3} \ln e$$Natural logarithm of$$e=1$$$$\frac{1}{3}\times1$$$$=\frac{1}{3}$$is this correct?

 

[BigBeachBanana]

Expanding $$\ln e^{1/3}$$$$\frac{1}{3} \ln e$$Natural logarithm of$$e=1$$$$\frac{1}{3}\times1$$$$=\frac{1}{3}$$is this correct?

Correct

 

[BigBeachBanana]

I have a similar question but it is in the expanded form precisely this(I wrote the final answer as well) $$\frac{3}{4} \ln e=\frac{3}{4}$$. So this is kinda similar to the "Power rule" but in reverse steps so the property used here is "Power rule" or it has some other name?

It's the same, you're just going from right to left, instead of left to right of

$\displaystyle \log_a(b^m) = m * \log_a(b)$

 

[pka]

For every number $\alpha\text{ , }~\log\left(e^{\alpha}\right)=\alpha$.

 

[JeffM]

For every number $\alpha\text{ , }~\log\left(e^{\alpha}\right)=\alpha$.

This is true only if we adopt the convention that $log(x) \text { MEANS } log_e(x)$.

Particularly in high school algebra, that convention is far from universal.

What are true universally are $ln(x) \text { MEANS } log_e(x)$ and $log_{\beta}(\beta^{\alpha}) = \alpha$.

 

[Dr.Peterson]

Question: Use the properties of natural logarithms to simplify the expression.
$$\ln e^{1/3}$$My answer: Inverse property should be used because
$$\ln e^{1/3}=\ln (1/e^3)$$
Is this correct? if not then which property should be used?

Ironically, you are exactly right that you should use the "inverse property". It's ironic because you don't seem to be using the term with the right meaning. (You seem to be thinking about reciprocals.)

The logarithm function is the inverse of the exponential function, which means that if you take the log of an exponential, you get back the input. This is what pka means by $\ln\left(e^{\alpha}\right)=\alpha$. It is also the third equation in the link you were given in #5.

What others are saying about using the power property and so on is not wrong, but this is the quick way, and very much worth learning.

 

[JasonTurner]

The statement is simple and asks the expression to be simplified. The equation uses the log system and is expressed in a fraction in the natural log. Using the basic laws of logarithm and fundamentals, the expression falls to $1/3.$
A natural logarithm solves the expression quickly, negating each log and putting the power values outside the record, giving you the answer without any hassle. There are several references listed in the discussion of this question.
The inverse property of logarithm means the input gets back when it is listed inside the log-in exponents. The same value returns on solving the log because it is the primary property of the natural logarithm. The value in exponents gets an out before it is applied on the constant holding the exponent.
Several references also highlight this logarithm property and are in the essential rule book of the logarithm. The connection highlights the derivatives equations of the logarithms and their exponentials.
These references highlight the derivation of this rule which states that any constant with the exponent in the log, the exponent will go out before the log is applied to that constant.