Thursday

January 29, 2015

January 29, 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

**Answer this Question**

**Related Questions**

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...