Please help me!!!! maths
posted by Seth on .
Solve for x: log5(x2)+log8(x4) = log6(x1). 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(x2)+log8(x4) = log6(x1)
log [40(x2)(x4)] = log 6(x1)
antilog it
40(x^2  6x + 8) = 6x6
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