Logarithms and such ...help?

Question

2^x = 3

(2.43) (10^x) = 1.84

Ln (x+5) = Ln (x-1) - Ln (x+1)

Answer 1

log(x^y) will give you y*log(x), same goes with ln.

The first one would therefore be:

log(2^x) = log(3)

<=>

x * log(2) = log(3)

<=>

x = log(3)/log(2) ~= 1.58

Answer 2

2^x = 3

log (2^x) = log 3  (take log of both sides)

x log 2 = log 3 (use property that log (a^r) = r log a)

x = log 3 / log 2 (divide both sides by log 2)

you could also say x = ln 3 / ln 2, same value

2.43 (10^x) = 1.84

10^x = (1.84 / 2.43)

log (10^x) = log (1.84 / 2.43)

because log (10^x) = x log 10 = x, this becomes

x = log (1.84 / 2.43)

ln (x + 5) = ln (x - 1) - ln (x + 1)

ln (x + 5) = ln [(x - 1) / (x + 1)]

now you have two logs with a common base, you can just equate the arguments

(x + 5) = (x - 1) / (x + 1)

(x + 5)(x + 1) = x - 1

x^2 + 6x + 5 = x - 1

x^2 + 5x + 6 = 0

(x + 3)(x + 2) = 0

x = -3 or x = -2

both of these make the arguments of the logs negative, so they are both extraneous

as written, there is no solution...

(which kind of makes sense... as x gets bigger and bigger, ln (x + 5) is going to get bigger and bigger

however, ln (x - 1) - ln (x + 1) is going to get closer to 0 (from the negative direction) because it will get closer to ln x - ln x ...