Question 597170
<pre>
The change of base formula is:

{{{log(B,U)}}} = {{{log(U)/log(B)}}}

or

{{{log(B,U)}}} = {{{ln(U)/ln(B)}}}
 
Therefore

{{{log(2,10)}}}= x = {{{log(2)/log(10)}}} ={{{.3010299957/1}}} = .3010299957   

or you can use:

{{{log(2,10)}}} = x = {{{ln(2)/ln(10)}}} ={{{.6931471806/2.302585093}}} = .3010299957 = .3010299957  

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Log12 (1243)=x

{{{log(12,1243)}}} = x = {{{log(1243)/log(12)}}} ={{{3.094471129/1.079181246}}} = 2.867424856   

or you can use:

{{{log(12,1243)}}} = x = {{{ln(1243)/ln(12)}}} ={{{7.125283092/2.48490665}}} = 2.867424856

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Notice that you may use either "log" (which is base 10) or "ln" (which is
base "e").  Both are on your calculator.  The numbers you divide will be
different but when you divide them you will always get the same answer either way.

Edwin</pre>