logx +7=1+log(x-1)
6=log(x-1)-logx
6=log[(x-1)/x]
10^6=(x-1)/x
1,000,000x=x-1
999,999x=-1
x=-1/999,999
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2x2+7/x1
The browser which is used for posting questions is almost totally useless for mathematical questions since it blocks most symbols.I am assuming that your question is about log base 3 of (x plus 1) plus log base 2 of (x-1).{log[(x + 1)^log2} + {log[(x - 1)^log3}/log(3^log2) where all the logs are to the same base - whichever you want. The denominator can also be written as log(3^log2)This can be simplified (?) to log{[(x + 1)^log2*(x - 1)^log3}/log(3^log2).As mentioned above, the expression can be to any base and so the expression becomesin base 2: log{[(x + 1)*(x - 1)^log3}/log(3) andin base 3: log{[(x + 1)^log2*(x - 1)}/log(2)
d/dx (2 log(1) + x) = 1
x=1
Y = 2x + 1