Why would anyone want to convert an exponential expression from one base to another? Why would logarithms help?

For that matter, what * are* exponents and logarithms?

If you simply want to learn the conversion formula, feel free to skip the introductions and link to the final page: **Finally: How to Convert the Base of an Exponent.**

## One Reason to Convert from one Exponential Base to Another

The chance of a tossed “fair” coin landing “heads” is 1/2. The chance of getting ten “heads” in ten tosses is “1/(2^10) = 1/1024”. That is not much worse than “1/1000 = 1/(10^3).”

Informally, that converted base-2 to base-10, and changed the exponent from ten to “approximately three.”

That’s easy enough if one selects a power of two that comes close to a power of ten, but surely there is a more accurate, general-purpose mathematical tool.

## Introduction to the Base and Exponent in Mathematics

Let’s answer that early question about “base”, “exponent” and “logarithm.”

Normally we use * base*-10 for everyday arithmetic. The ten digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} indicate specific values. Writing a number such as “twenty-seven and a half = 27.5” uses three of those digits. The value is found by using powers of ten.

“27.5 = 2*10 + 7*1 + 5/10 = 2*(10^1) + 7*(10^0) + 5*(10^-1)” is a very explicit way to show this value. The ‘*’ is the multiplication operator, and the ‘^’ is the * exponentiation* operator.

The * exponent* shows the number of times the

**base**is multiplied by itself. For any non-zero real number ‘b’ as the base and ‘j’ as a positive integer, “b^j = b*b*…*b” where the ‘b’ appears ‘j’ times. So “b^1 = b”.

By definition, “b^0 = 1” for any non-zero, real number ‘b’. Some would define “0^0 = 1” also, but others leave it undefined.

An exponent can be a negative integer. If ‘j’ is a positive integer, then “b^(-j) = 1/(b^j)”.

An exponent can be a fraction, or “rational” number. If ‘j’ and ‘k’ are positive integers, “b^(1/k) = the k-th root of b”. From the coin-toss example, “2 is the 10th root of 1,024.”

In other words, “b = ( b^(1/k) )^k”. Multiply the k-th root of ‘b’ by itself ‘k’ times, and you have the value ‘b’.

Then “b^(j/k) = the k-th root of b^j”.

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.