Asked by Jeconiah
Solve for x: log5(x-2)+ log5 (x-4) = log6(x-1)
Answers
Answered by
Reiny
log<sub>5</sub> ( (x-2)(x-4) ) = log<sub>6</sub> (x-1)
clearly x > 4
log (x^2 - 6x + 8) /log5 = log(x-1)/log6
log6 log(x^2 - 6x + 8) = log5 log(x-1)
log (x^2 - 6x + 8)^(log6) = log (x-1)^(log5)
anti-log it
(x^2 - 6x + 8)^log6 = (x-1)^log5
..... oh my, did you really mean for those bases to be diffferent ??
I ran it through Wolfram and got
x = 5.1409 , rejecting the smaller.
http://www.wolframalpha.com/input/?i=log%28+%28x-2%29%28x-4%29+%29%2Flog%285%29+%3D+log%28x-1%29%2Flog%286%29
clearly x > 4
log (x^2 - 6x + 8) /log5 = log(x-1)/log6
log6 log(x^2 - 6x + 8) = log5 log(x-1)
log (x^2 - 6x + 8)^(log6) = log (x-1)^(log5)
anti-log it
(x^2 - 6x + 8)^log6 = (x-1)^log5
..... oh my, did you really mean for those bases to be diffferent ??
I ran it through Wolfram and got
x = 5.1409 , rejecting the smaller.
http://www.wolframalpha.com/input/?i=log%28+%28x-2%29%28x-4%29+%29%2Flog%285%29+%3D+log%28x-1%29%2Flog%286%29
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