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Solve: log(6) (x^2-5x) = 1
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that's log base 6 just to clarify
\[\log_b(\xi)=1\iff \xi =b\]
so \[\log_6(\xi)=1\iff \xi =6\]
ooohhhh I know what your saayin now
ok and then what after that
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right you only job is to solve \[x^2-5x=6\]
wait now that doesn't make sense
I thought they have to be same base?
all I did was switch those and then I could just dropo the log?
i will let you figure out why if \[\log_6(x^2-5x)=1\] then \[x^2-5x=6^1\]
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heres what ihave after I switched them |dw:1368478776006:dw|
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