 | Charles Davies - 1841 - 414 páginas
...member, we have but since a is the base of the system, m+n is the logarithm JJ/xJV; hence, The sum of the logarithms of any two numbers is equal to the logarithm of their product. Therefore, the addition of logarithms corresponds to the multiplication of their... | |
 | Charles Davies - 1842 - 284 páginas
...logarithms of any two numbers equal ? To what then, will the addition of logarithms) correspond ? The sum of the logarithms of any two numbers is equal to the logarithm of their product. Therefore, the addition of logarithms corresponds to the multiplication of their... | |
 | James Thomson - 1848 - 326 páginas
...logarithms of numbers are other numbers depending on them, and characterized by the property, that the sum of the logarithms of any two numbers is equal to the logarithm of their product. Thus, log 6+log c=log (6c). Hence also, since b=-.c, it follows, that c log6=log-+logc;... | |
 | Charles Davies - 1848 - 300 páginas
...logarithms of any two numbers equal ? To what then, will the addition of logarithms correspond ? The sum of the logarithms of any two numbers is equal to the logarithm of their product. Therefore, the addition of logarithms corresponds to the multiplication of their... | |
 | John Radford Young - 1851 - 266 páginas
...we shall see when a few obvious propositions in the theory of logarithms are stated. 1 1 7. Tne sum of the logarithms of any two numbers is equal to the logarithm of their product. Let a* = n, and a'—n' .: aI+•'=nn'; therefore, if a be the base of the system... | |
 | Adrien Marie Legendre - 1852 - 436 páginas
...Multiplying equations (1) and (2), member by member, we have, or, m + n=log (Mx N); hence, The sum of the logarithms of any two numbers is equal to the logarithm of their product. 4. Dividing equation (1) by equation (2), member by member, we have, mn MM 10 -=_r~0r,... | |
 | Charles Davies - 1886 - 340 páginas
...equations (1) and (2), member by member, we have lO"""" = MxN or, m+n — log MxN : hence, The sum of the logarithms of any two numbers is equal to the logarithm of their productDividing equation (1) by equation (2), member by member, we have " ,m— n M ' M 10... | |
 | Charles Davies - 1854 - 436 páginas
...equations (1) and (2), member by member, we have, 10m+ n = Mx N or,m + n=log (Mx N) ; hence, The sum of the logarithms of any two numbers is equal to the logarithm of their product. 4. Dividing equation (1) by equation (2), member by member, we have, JO™ »BB_OTjW_Wesi0g—... | |
 | Charles Davies - 1854 - 446 páginas
...Multiplying equations (1) and (2), member by member, we have, Wm + n = MxN OT,m + n=log(MxN); hence, The sum of the logarithms of any two numbers is equal to the logarithm of their product. 4. Dividing equation (1) by equation (2), member by member, we have, 10m~n = -^or,... | |
 | Adrien Marie Legendre, Charles Davies - 1857 - 442 páginas
...= Jf (1) 10" = ^ (2). Multiplying equations (1) and (2), member by member, we have, hence, The sum of the logarithms of any two numbers is equal to the logarithm of their product. 4. Dividing equation (1) by equation (2), member by member, we have, , , Jf J/ 10m~"... | |
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JJ/xJV; hence, The sum of the logarithms of any two numbers is equal to the logarithm of their product. Therefore, the addition of logarithms corresponds to the multiplication of their numbers. 
