Question

solve for x: log(3x-7)-log(x-2)=1

Answer 1

might be a good idea to take the base of each side ...

Answer 2

if log pertains to base 10, then 10 each side

in higher math courses there is only 1 log, the natural log and therefore your notation would refer to a base e

Answer 3

the idea is:\[b^{n+m}=b^n~b^m\]and \[b^1=b\]
along with:\[b^{(k)_b}=k\]

Answer 4

will it look like this 10^log(3x-7) - 10^log(x-2) = 1
3x-7-x-2=100
x=55 ?

Answer 5

good try, but not quite:

10^ [log(3x-7) - log(x-2)] = 10^1

10^[log(3x-7)] * 10^[log(x-2)] = 10

(3x-7)(x-2) = 10

Answer 6

even i messed that up lol, i dropped the negative

Answer 7

10^ [log(3x-7) - log(x-2)] = 10^1

10^[log(3x-7)] * 10^[-log(x-2)] = 10
^^

10^[log(3x-7)] * 10^[log(1/(x-2))] = 10

(3x-7)/(x-2) = 10

thats better

Answer 8

so the question becomes:

3x-2 = 10(x-2), solve for x

Answer 9

gotta read thru the typos this morning .... 3x-7 = 10(x-2)

Answer 10

why is it [-log(1/x-2)] ?

Answer 11

by this property, and it went from -log(a) to log(1/a)

\[k~log(a)=log(a^k)\]
since k=-1, thats another way of presenting the reciprocal of a since:\[a^{-1}=1/a\]

Answer 12

another way to approach it is this, same results tho: using the log property that:\[log(a/b)=log(a)-log(b)\]
we can combine the logs, then base it

Answer 13

that's exponential property or power property

Answer 14

oh okay :)

Answer 15

:) i cant recall what they name the things

Answer 16

x= 13/7

Answer 17

thats what i get

Answer 18

http://www.wolframalpha.com/input/?i=log%283x-7%29-log%28x-2%29%3D1&a=*FunClash.log-_*Log10-

the wolf agrees, assuming log refers to base10