posted by Seth

Solve for the value of x: log5(x-2)+log8(x-4)=Log6(x-1)..For your information that 5,8 and 6 are bases not just multipliers. Thank you.

1. Steve

log 5(x-2) + log 8(x-4) = log 6(x-1)
log [5(x-2)*8(x-4)] = log 6(x-1)

so, if the logs are =, so are the expressions, and we have

40(x-2)(x-4) = 6(x-1)
20(x-2)(x-4) - 3(x-1) = 0
20x^2 - 123x + 163 = 0

now just use the quadratic formula to see the solutions. Check to be sure the values are defined in the original expressions.

2. Seth

6log(x^2+1)-x=0

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