Question

Hi Guys.
Solve for x.
2logx=log2+log(6x)-10)
Need a lower and upper value....drawing a blank on where to begin.

Answer 1

i can't really read it
is it
\[2\log(x)=\log(2)+\log(6x-10)\]?

Answer 2

2 log x = log 2 + log(6x − 10)
need an upper and lower value.
I thought to start I should move the 2 log x to the other side but I don't know what then.

Answer 3

that is probably the best idea, get
\[2\log(x)-\log(6x-10)=\log(2)\] then combine the stuff on the left in to a single log using
\[n\log(x)=\log(x^n)\] and
\[\log(A)-\log(B)=\log\left(\frac{A}{B}\right)\]

Answer 4

i'll try it. one moment

Answer 5

right on the money....it became x^2-12x+20....values on 2 and 20.
I may have one or two more I could use a bit of direction on. mind if I look u up later? thanks

Answer 6

ok.....log2(x + 8) − log2(x − 8) = 3
initially I am thinking make everything base 10 to get
2(x+8)-2(x-8)=3. Am I on the right track?

Answer 7

leave it as base 2

Answer 8

combine in a single log like before

Answer 9

log 2 ((x+8)/(x-8))=3?