 | Charles Davies - 1841 - 414 sider
...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 sider
...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 sider
...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 sider
...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 sider
...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 sider
...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 sider
...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 sider
...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 sider
...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 sider
...= 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. 
