exponent

Feb 2010
50
0
hi,

if we have: 2^10 = 1024

what is 10^? = 1024

Since this is a small number I can do rote trying and come to 10^3=1000 ~1024 but how I can approximate this with some formula or some algorithms when I have really big number.

For example to convert from base 2 to 10 and play with the exponents to get some number?

Thanks!
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
hi,

if we have: 2^10 = 1024

what is 10^? = 1024

Since this is a small number I can do rote trying and come to 10^3=1000 ~1024 but how I can approximate this with some formula or some algorithms when I have really big number.

For example to convert from base 2 to 10 and play with the exponents to get some number?

Thanks!
\(\displaystyle 10^x = 2^{10}\)

log (base 10) both sides ...

\(\displaystyle \log(10^x) = \log(2^{10})
\)

power property of logs ...

\(\displaystyle x\log{10} = 10\log{2}
\)

\(\displaystyle x = 10\log{2}\)

note that \(\displaystyle \log{2} \approx 0.30103\)
 
Jun 2008
292
87
xlog_2[10]=log_2[1024]
x=log_10[2]*10
note: log_a=1/log_b[a]
 
Feb 2010
50
0
\(\displaystyle 10^x = 2^{10}\)

log (base 10) both sides ...

\(\displaystyle \log(10^x) = \log(2^{10})
\)

power property of logs ...

\(\displaystyle x\log{10} = 10\log{2}
\)

\(\displaystyle x = 10\log{2}\)

note that \(\displaystyle \log{2} \approx 0.30103\)
But what if I have the opposite?

10^3=2^x or 10^3=2^?

How I can solve this case?

And what if I would like to manipulate other exponents not just 2 to 10 or 10 to 2?

Any help...?

Thanks!
 

e^(i*pi)

MHF Hall of Honor
Feb 2009
3,053
1,333
West Midlands, England
But what if I have the opposite?

10^3=2^x or 10^3=2^?

How I can solve this case?

And what if I would like to manipulate other exponents not just 2 to 10 or 10 to 2?

Any help...?

Thanks!
You can use any base you like and then use algebra to simplify.

For example I will solve \(\displaystyle 10^3 = 2^x\) using base \(\displaystyle e\)

\(\displaystyle 3\ln(10) = x\ln(2)\)

\(\displaystyle x = \frac{3\ln(10)}{\ln(2)}\)


More generally if a^b = c^x where a,b and c are positive constants and \(\displaystyle a \neq c\) then \(\displaystyle x = b\log_c(a)\) or \(\displaystyle x = \frac{b\log_e(a)}{\log_e(c)}\)
 
Last edited:

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
But what if I have the opposite?

10^3=2^x or 10^3=2^?

How I can solve this case?

And what if I would like to manipulate other exponents not just 2 to 10 or 10 to 2?

Any help...?

Thanks!
same procedure for the general case ...

\(\displaystyle a^b = b^x\)

use either a base 10 or base e log because those values are readily available from any scientific calculator ...

\(\displaystyle \log(a^b) = \log(b^x)\)

\(\displaystyle b\log{a} = x\log{b}\)

\(\displaystyle x = \frac{b\log{a}}{\log{b}}\)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Note that in neither of those cases, \(\displaystyle 2^{10}= 1024\) so \(\displaystyle 10^?= 1024\) nor \(\displaystyle 10^3= 2^?\) can you expect the answer to be an integer. That is because 2 to any power can have only 2 as a prime factor while 10 to a power greater than 0 will have prime factors of both 2 and 5.
 
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