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
Converting base-2 to base-1o for a coin toss : Image by ICMA Photos
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”.
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.