# Thread: How do you evaluate an expression with "e" as the base?

1. ## How do you evaluate an expression with "e" as the base?

so something like e^3.12 or e^1/5

2. ## Re: How do you evaluate an expression with "e" as the base?

Originally Posted by vk985
so something like e^3.12 or e^1/5

A calculator would solve that quickly for you since it's a constant in the exponent.

If it were a variable we'd use the natural log: $y = e^x \Leftrightarrow \ln(y) = x$

If we wanted the exact answer we'd leave it like that

3. ## Re: How do you evaluate an expression with "e" as the base?

To evaluate $e^{3.2}$, using the calculator that comes with Windows, enter "3.2", then click on the "inv" button. The "ln" button changes to "e^x". Pressing that gives "24.532530197109348643560263727964".

To evaluate $e^{1/5}$, much the same. Enter "1", "/", "5", "=" and you will see "0.2". Now, click on "inv" and then "e^x".
The result is "1.2214027581601698339210719946397".

Some calculators have both " $e^x$" and " $ln(x)$" keys. The point is, of course, that $ln(x)$ and $e^x$ are inverse functions: $e^{ln(x)}= ln(e^x}= x$.

In the years "BC" (before calulators) people would look up logarithms and exponentials in tables. Books would have tables of, typically, logarithms, and to find the exponential you would use the logarithm table "in reverse".

There is no simple way to "calculate" a logarithm or exponential although you could use the Taylor's series to approximate: $e^x= 1+ x+ \frac{1}{2}x^2+ \frac{1}{6}x^3+ \cdot\cdot\cdot+ \frac{1}{n!}x^n+ \cdot\cdot\cdot$. An approximation because that is an infinite series and you can only calculate a finite number of terms. For example, with x= 3.2, the first four terms give $1+ 3.2+ \frac{1}{2}(3.2)^2+ \frac{1}{6}(3.2)^3= 1+ 3.2+ 5.12+ 5.46+ 4.37+ 2.79= 21.94$, not yet close to 24.....