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March 31, 2015

March 31, 2015

Posted by **Seth** on Wednesday, April 3, 2013 at 5:33am.

- Please help me!!!! maths -
**Reiny**, Wednesday, April 3, 2013 at 12:04pmI 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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