# exponent

• May 31st 2010, 06:48 AM
exponent
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!
• May 31st 2010, 06:56 AM
skeeter
Quote:

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 $\displaystyle \log{2} \approx 0.30103$
• May 31st 2010, 07:00 AM
nikhil
xlog_2[10]=log_2[1024]
x=log_10[2]*10
note: log_a[b]=1/log_b[a]
• May 31st 2010, 07:17 AM
Quote:

Originally Posted by skeeter
$\displaystyle 10^x = 2^{10}$

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!
• May 31st 2010, 07:41 AM
e^(i*pi)
Quote:

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 $\displaystyle x = \frac{b\log_e(a)}{\log_e(c)}$
• May 31st 2010, 07:44 AM
skeeter
Quote:

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 ...

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

$\displaystyle b\log{a} = x\log{b}$

$\displaystyle x = \frac{b\log{a}}{\log{b}}$
• May 31st 2010, 08:25 AM
HallsofIvy
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.