MHB What is the difference between ln 4 and log 4?

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The discussion clarifies the difference between ln 4 and log 4, emphasizing that ln 4 refers specifically to the natural logarithm (log base e), while log 4 can denote logarithms of different bases, often base 10. The integral of (1/x) from 1 to 4 correctly evaluates to ln 4, as ln(x) is the antiderivative of (1/x). The confusion arises from varying conventions in notation, where ln(x) is consistently used for log base e, while log(x) may refer to either log base 10 or log base e depending on the context. Ultimately, ln 4 is greater than log 4 if log 4 is interpreted as log base 10. Understanding the notation is crucial for accurate mathematical communication.
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I am reviewing Calculus 1 integration learned long ago in the 1990s.

Integrate (1/x) dx from 1 to 4.

The textbook answer is ln 4.

However, many of my friends tell me that the answer can also be written as log 4.

But, ln 4 does NOT equal log 4.

In fact, ln 4 > log 4.

Who is right? Why?
 
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It depends on what shorthand notation one uses for $\log_e(x)$.
 
MarkFL said:
It depends on what shorthand notation one uses for $\log_e(x)$.

The integral of (1/x) is ln x.

From 1 to 4:

ln 4 - ln 1

ln 4 - 0

Answer: ln 4

However, my friend said the following:

log 4 - log 1

log 4 - 0

log 4

Who is right?
 
I'll expand a bit beyond MarkFL's comment. You problem is a matter of notation, not a conceptual one.

In many sources we have that [math]ln(x) = log_e(x)[/math]. Sometimes you will see [math]log_e(x) = log(x)[/math]. In fact some also say that [math]log_{10}(x) = log(x)[/math].

In your case we need the [math]ln(4) = log_e(x)[/math].

-Dan
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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