# How do we calculate log_3 10?

• Jan 29th 2013, 01:28 AM
kpkkpk
How do we calculate log_3 10?
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?(Nerd)
• Jan 29th 2013, 05:24 AM
Prove It
Re: How do we calculate log_3 10?
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).
• Jan 30th 2013, 01:50 AM
kpkkpk
Re: How do we calculate log_3 10?
Quote:

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).