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Solve for x: log5(x-2)+log8(x-4) = log6(x-1). I do not have any idea on this topic.

I am not sure if the 5 , 8 , and 6 are bases of the logs
or just multipliers.
Let's hope they are just multipliers, or else it would be a terrible terrible mess

by the laws of logs
log5(x-2)+log8(x-4) = log6(x-1)
log [40(x-2)(x-4)] = log 6(x-1)
anti-log it

40(x^2 - 6x + 8) = 6x-6
40x^2 - 240x + 320 - 6x + 6 = 0
40x^2 - 246x + 326 = 0
20x^2 - 123x + 163 = 0
by the formula
x = (123 ± √2089)/40
= 4.2176 or 1.9324 , but from the original equation , x > 4

so x = 4.2176

I tested the answer, it works

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