# 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!

#### skeeter

MHF Helper
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 $$\displaystyle \log{2} \approx 0.30103$$

#### nikhil

xlog_2=log_2
x=log_10*10
note: log_a=1/log_b[a]

$$\displaystyle 10^x = 2^{10}$$

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
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 $$\displaystyle x = \frac{b\log_e(a)}{\log_e(c)}$$

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#### skeeter

MHF Helper
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
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.

• Rapha