Final question and then I will give my text book a try again: write as a single log: log(2x+1)-(1/3)log(x+1)=?

Answer 1

log(2x+1)-(1/3)log(x+1)
=log(2x+1) - log( (x+1)^(1/3) )
=log( (2x+1) / (x+1)^(1/3) )

Answer 2

2x+1)-(1/3)log(x+1)=?

log(2x+1)-(1/3)log(x+1)
=log(2x+1) - log( (x+1)^(1/3) )
=log( (2x+1) / (x+1)^(1/3) )


nverse relations

The inverse of any exponential function is a logarithmic function. For, in any base b:

i) blogbx = x,

and

ii) logbbx = x.

Rule i) embodies the definition of a logarithm: logbx is the exponent to which b must be raised -- to produce x.

Rule ii) we have seen before (Topic 20).

Now, let

f(x) = bx and g(x) = logbx.

Then Rule i) is f(g(x)) = x.

And Rule ii) is g(f(x)) = x.

This satisfies the definition (Topic 19) of a pair of inverse functions. Therefore for any base b, the functions

f(x) = bx and g(x) = logbx

are inverses.

Problem 3. Evaluate the following.

a) log225 = 5 b) log 106.2 = 6.2 c) ln ex + 1 = x + 1

d) 2log25 = 5 e) 10log 100 = 100 f) eln (x − 5) = x − 5

Answer 3

log (2x+1) - (1/3)log(x+1)
= log(2x+1) - log (x+1)^(1/3)
=log ((2x+1) / (x+1)^(1/3))



log ((2x+1) / (x+1)^(1/3)) is the ans

Answer 4

log ( (2x+1)/ ((x+1)^1/3))