## Who Were Euler and Napier?

Leonhard Euler was a Swiss mathematician who lived from 1707 to 1783.

One of his many contributions to mathematics was the proof that ‘e’ is an irrational number.

Euler’s formula for ‘e’ is “e^(i*x) = cosine(x) + i*sine(x)”, where ‘i’ is the square root of negative one, and ‘x’ is any real number.

Although ‘e’ may be named for the word “* e*xponential” or in his honour as “

*uler’s number”, he also has the completely different “γ (gamma) = Euler’s constant = 0.57721…”.*

**E**John Napier “invented” logarithms, and was particularly involved with natural logarithms.

Born in 1550, this Scottish landowner and polymath made other significant contributions to mathematics. He died in 1617, but several of his books were published posthumously.

By the way, he defined the “* Naperian logarithm*” which

*differs from*the natural logarithm. His definition for ‘L’, the Naperian logarithm of a number ‘N’, is:

- “N = 10^7 * (1 – 10^(-7) )^L”

Enjoy solving for ‘L’.

## Finally, How to Convert the Base of an Exponent

Here is the general rule for converting an exponential expression from one base to another.

Let ‘b’ be the first base, and ‘a’ the target base. Let ‘m’ be the known exponent, and ‘x’ is the unknown.

To solve for ‘x’ when “b^m = a^x”, use

- “
“.**x = m*ln(b)/ln(a)**

If you plan to convert between the same bases, it’s worthwhile to calculate the conversion factor, “ln(b)/ln(a)”, and keep it for reference.

* References*:

Sondow, Jonathan and Weisstein, Eric W. *e*. (2012). MathWorld–A Wolfram Web Resource. Accessed October 7, 2012.

Weisstein, Eric W.* Base; Logarithm; Natural Logarithm; Napierian Logarithm. *(2012).* *MathWorld — A Wolfram Web Resource. Accessed October 7, 2012.

John says

Re OP:

same idea, easier for me to understand:

1. “b^m = a^x”, to solve for ‘x’.

2. “log[a}(b^m) = x”. take log[a] of both sides.

3. “m * log[a}(b) = x”. pull the m out of the log

4. “m * ln(b)/ln(a) = x”. replace with natural logs

Re Chuck:

from what you described:

b^m = a^x

b/a = c

c * m =? x

c should be 0.301 from OP’s calculations rather than 5

also log[2]( 10^15 ) = 49.829 not 75

in other words:

10^14 = 2^46.507

10^15 = 2^49.829

Chuck Crosthwait says

I’m pretty sure I found an easier formula for converting an exponent series from base10 to base2. I’ll try to express it here with a known value.

10^14=2^x

Using the formula above you’d use

x=14*ln(10)/ln(2)

and arrive at 70.

If you use some funky cross multiplication I haven’t seen anywhere you could do the following instead:

10^14=2^x

10/2=5

5*14=70

10^14=2^70

I’d like to preface with the fact that I did not locate this formula anywhere after searching for ~1hr. I tried to remember the general formula for finding exponent X in a known system. I have only tried this converting from base10 to base2 and back and tried with other known values, e.g.:

10^15=2^x

10/2=5

5*15=75

10^15=2^75

and this is borne out in the formula

x = m*ln(b)/ln(a)

for finding

b^m = a^x

so it works. Did I stumble upon a formula I haven’t found before or is this something new and just too simple to have been considered before?

Please do reply, I really want to know if there’s some unknown mathematical principle I needed to know or knew and forgot.