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Andrew123
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Why does e ^ (lnx) = x? where ln is log base e. Why is the integral of 1/x ln(x) and not log(x) ie log base 10 x? Is there some good information somewhere explainined this please? Cheers
Andrew123 said:Why does e ^ (lnx) = x?
Why is the integral of 1/x ln(x)
Werg22 said:As Integral pointed out, the relationship
[tex]
\int _1 ^e \frac {dx} x = 1
[/tex]
is an important one in elucidating this question. Usually, we define log x as
[tex]
\int _1 ^x \frac {dt} t
[/tex]
The constant e can be defined as the limit of (1 + 1/n)^n and it can be shown that log e = 1. That said, the equation log (a^b) = b * log a can be derived purely from the definition of log as an integral*. Then follows that log (e^x) = x*log e = x. This process shows that e^x is the inverse function of log x. And so e^log x = x **
* Technically, I know only of such a development that assumes b is rational and expands to all real numbers later in the program, but assuming this is valid for all real b's still conveys the main idea.
Natural logs, or ln, are the inverse of the exponential function e^x. This means that taking the natural log of both sides of an equation involving e allows us to isolate the variable and solve for it.
Yes, natural logs can be used to solve any equation that involves the variable being raised to a power of e. However, they may not always be the most efficient method of solving the equation.
Yes, there are a few rules to keep in mind when using natural logs to solve equations involving e. These include the property ln(ab) = ln(a) + ln(b) and the rule for solving ln(e^x) = x.
Yes, natural logs can be used to solve equations with any base. However, the equation must be rewritten in terms of e first. For example, to solve an equation with base 10, we would use the rule ln(x) = ln(10^x).
No, it is not always necessary to use natural logs to solve equations involving e. Other methods, such as substitution or algebraic manipulation, may be more efficient in certain cases.