How do we calculate log_3 10?

**kpkkpk** · January 29th 2013, 01:28 AM

2^7=2*2*2*2*2*2*2=128 poses no practical problem.
But 3^4567 does!
From my previous question (18th Jan) and kind answer I have learned that 3^4567=10^x, and from that equation we can deduce that x=4568/log_3 10.
If we insert these values to calculator, we can find that answer is about 4568/2,095...=2180,4296... which means that there are 2180 integers in an answer if we "open" 3^4567.

But where does this denominator value log_3 10 comes from?
What kind of a number it and other log values are?
Are they always decimal numbers? ...irrational numbers? ...transcendental numbers?

As far as I know Mr. Napier (who invented logarithms) and his contemporaries had to use a lot of time to find approximate values for logarithms they used...and even so, they only tabulated some 5 decimals for each case.
So, what kind is an algorithm that creates logaritms nowadays?

**Prove It** · January 29th 2013, 05:24 AM

For starters, you need to remember the logarithm law $\displaystyle \begin{align*} \log_b{ \left( m^p \right) } = p\log_b{(m)} \end{align*}$ . So if you start with your equation $\displaystyle \begin{align*} 3^{4567} = 10^x \end{align*}$ and take a logarithm of ANY base to both sides (in this case they chose 3 because it cancels easiest on the left hand side), we have

$\displaystyle \begin{align*} 3^{4567} &= 10^x \\ \log_3{\left( 3^{4567} \right) } &= \log_3{\left( 10^x \right) } \\ 4567 &= \log_3{ \left( 10^x \right) } \\ 4567 &= x \log_3{(10)} \textrm{ from the logarithm law mentioned earlier} \\ \frac{4567}{\log_3{(10)}} &= x \end{align*}$

It would have been just as easy to cancel on the RHS by taking both sides to a base 10 logarithm, or even taking both sides to any other logarithm as well.

Most logarithm values are transcendental, unless the number you are taking a logarithm of is a power of the same base of your logarithm, because then you have natural cancellation. If your logarithm is transcendental though, decimal approximations can be made using a Taylor Series (which calculators would be programmed with).

**kpkkpk** · January 30th 2013, 01:50 AM

Originally Posted by Prove It

For starters, you need to remember the logarithm law $\displaystyle \begin{align*} \log_b{ \left( m^p \right) } = p\log_b{(m)} \end{align*}$ . So if you start with your equation $\displaystyle \begin{align*} 3^{4567} = 10^x \end{align*}$ and take a logarithm of ANY base to both sides (in this case they chose 3 because it cancels easiest on the left hand side), we have

$\displaystyle \begin{align*} 3^{4567} &= 10^x \\ \log_3{\left( 3^{4567} \right) } &= \log_3{\left( 10^x \right) } \\ 4567 &= \log_3{ \left( 10^x \right) } \\ 4567 &= x \log_3{(10)} \textrm{ from the logarithm law mentioned earlier} \\ \frac{4567}{\log_3{(10)}} &= x \end{align*}$

It would have been just as easy to cancel on the RHS by taking both sides to a base 10 logarithm, or even taking both sides to any other logarithm as well.

Most logarithm values are transcendental, unless the number you are taking a logarithm of is a power of the same base of your logarithm, because then you have natural cancellation. If your logarithm is transcendental though, decimal approximations can be made using a Taylor Series (which calculators would be programmed with).

OK. Thanks for your answer, Prove it.
So, by using logarithms help me to find answer to question: "how many integers there are if we "open" 3^4567. But, compared to raw repetitive multiplications, do they have any help if I try to find what these integers are precisely?
So, is it more economical to use Taylor series to find enough many decimals in order to find precise values for large raisings or is it just waste of time to use them compared to raw repetitive multiplications? Or is there any 3rd way how we (=computers/calculators) in practice "open" these integer raisings?

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