# Help with deriving relationships starting with the identity a^x = e^xlna

• CTil
In summary, the conversation discusses deriving relationships between (a) e^x and 10^x and (b) ln x and log x using the identity a^x = e^xlna. The first part involves rewriting 10^x in the form of e^b and the second part involves connecting expressions with x and replacing N with x.
CTil
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Hi there - it has been quite a long time since I took Calculus. I am trying to brush up and understand where to start with this question:

Starting with the identity a^x = e^xlna, derive the relationships between (a) e^x and 10^x; (b) ln x and log x. Note: log x = log10 x unless otherwise specified.

I know how the a^x was derived, but I'm honestly not sure what a) is asking? Is the 10^x regarding logarithm?
I have re-written the equation to a^x = (e^ln(a))^x

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Can you write ## 10 ## in the form ## e^b ##? What is ## b ##? That should help get you started.

here's where I am at...

f(x) = 10^x; 10^x = e^xln10; f′(x) = e^xln10 (ln 10) = 10^x ln 10
then from there I got 10^x 2.303 log 10...

The second part (b), in my opinion, is kind of a clumsy one in using the equation that is given. If you let ## N=a^x=e^{x \ln{a}} ##, then you can write, with ## a=10 ##, that ## x=\log_{10}(N) ## , but also ## \ln{N}=x \ln{10} ##. (Without introducing ## N ##, I think it is more difficult). Connect these last two expressions that each have an ## x ##. ## \\## Finally replace ## N ## with an ## x ##. (## N ## represents an arbitrary number in this last equation that you obtain, so you can replace it with any letter you choose. This new ## x ## is, of course, totally unrelated to the first ## x ##).

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So if I am following you correctly, ln N/ln 10 = log10 (N) or ln X / ln 10 = log 10 X ?

CTil said:
So if I am following you correctly, ln N/ln 10 = log10 (N) or ln X / ln 10 = log 10 X ?
Correct.

## 1. How can I derive relationships starting with the identity a^x = e^xlna?

In order to derive relationships starting with the identity a^x = e^xlna, you can use logarithms to simplify the equation. Taking the natural log of both sides will result in ln(a^x) = ln(e^xlna). Then, using the power rule of logarithms, you can rewrite the equation as xlna = xlna. This shows that the original identity is true and can be used to derive further relationships.

## 2. Can I use any value for "a" in the identity a^x = e^xlna?

Yes, you can use any positive value for "a" in the identity a^x = e^xlna. This includes whole numbers, fractions, and irrational numbers. However, keep in mind that the value of "a" will affect the resulting relationships that you derive.

## 3. What types of relationships can be derived using the identity a^x = e^xlna?

The identity a^x = e^xlna can be used to derive exponential and logarithmic relationships. This is because the identity connects the exponential function a^x with the natural logarithm function ln(a). By manipulating the identity, you can create new equations that demonstrate the relationship between these two functions.

## 4. Are there any limitations to using the identity a^x = e^xlna for deriving relationships?

One limitation to using the identity a^x = e^xlna is that it only applies to positive values of "a". Additionally, the identity only works for real numbers. It cannot be used with complex numbers. Furthermore, the resulting relationships may only be applicable within certain domains and ranges, depending on the values of "a" and "x" used.

## 5. How can I use the derived relationships from the identity a^x = e^xlna in practical applications?

The derived relationships from the identity a^x = e^xlna can be applied in many fields of science and mathematics. For example, they can be used in finance and economics to model compound interest, in physics to describe exponential decay or growth, and in biology to understand population growth. They can also be used in data analysis and modeling to make predictions and projections.

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