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log_10(6 x-1) - log_10(x-1) = 1
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Are you just putting up all your logarithm homework?
no these are the ones i don't understand how to do
log(x)-log(y)=log(x/y) as shown previously. 10^log_10(x/y)=10^whatever's on this side. Solve that way.
\[\log_{10}(6x-1)-\log_{10}(x-1)=1\] \[\large10^{\log_{10}(6x-1)}-10^{\log_{10}(x-1)}=10^1\] \[(6x-1)-(x-1)=10\] \[6x-1-x+1=10\] \[5x=10\] \[x=\frac{10}{5}\] \[x=2\]
okay so whenever a multiplication you do subtraction? I thought that was for addition and subtraction was division.
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