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)
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
Please Reiny Solve it For me Again! please...... - Solve for the value of x: ...
Maths - Solve for x: log5(x-2)+ log5 (x-4) = log6(x-1)
Math - The problem I have to solve is log with base 2 ^6 multiply by log base 6...
Algebra II - Log6(6^9)=??? Answer: 9 Ine^-3x=??? Answer: -3x Use log5(2) = 0....
College Algebra - I REALLY don't understand the reason/basis/use of logarithms. ...
Math - How do I solve: log6 (x^2-5x) = 1 1/2log5(r-2)-log5 r) log2 (x-1/x^8) I ...
Math - log(x-2) +log8 (x-4) =log6 (x-1)
math 12 - log6=x and log8=y, determine an expression for log3 in terms of x an y...
precalculus - solve the logarithmic equation . express solution in exact form ...
Algebra 2 - Can someone please explain to me how to solve logarithmic equations...