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log5(x-1) + log5(x-2) - log5(x+6) = 0

I feel like this is really simple but I cannot get the answer.

This is what I attempted

Since they have the same base, I multiplied the top

log5[(x-1)(x-2)] - log5(x+6) = 0

This gives me

log5(x2-3x+2) - log5(x+6) = 0

At this point I did not now what to do, I tried 2 methods,

Since they are the same base again, I divided the top this time

log5(x2-3x+2) / (x+6) = 0

but I don't know how to continue


I moved - log5(x+6) to the other side

log5(x2-3x+2) = log5(x+6)

Since they have the same base I ignored them so

x2-3x+2 = x+6

Which gave me

x2 - 4x - 4

Which cannot be factored.

Please help, thankyou smile.png

Edited by Acnhduy
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[(x-1)(x-2)]/(x+6) = 0
That is not correctly derived. Exp(0) = 1, not 0. The OP has it right, in the part where they "ignored the base".



For many people, btw, the quadratic formula is more easily remembered by sound - memorized as a one would a short poem or limerick - rather than by sight. If you are one of those people, say it as you are writing it - by hand, with a pencil - a few times, and it will stick.

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[math]\log_5 (x-1)+\log_5 (x-2)-\log_5 (x+6)=0[/math]


There are probably more efficient ways, but the first thing I thought of doing was...


[math]5^{\log_5 (x-1)+\log_5 (x-2)-\log_5 (x+6)}=5^0[/math]


Take 5 to the power of both sides. After some simplification...




From there, you can bring out the denominator to the other side and rearrange the equation into standard quadratic form.

Edited by Amaton
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