Question

Express log x – 2 log x + 3 log (x+1) – log (x2 - 1) as a single logarithm​

Answer 1

ANSWER:

=−logx+3log(x+1)−log(x^2−1)=

=−log(x)+log((x+1)3)−log(x^2−1)=

=log(x+1)^3/x−log(x^2−1)=

=log(x+1)^3/x(x^2−1)=

=log(x+1)^3/x(x−1)(x+1)=

=log(x+1)^2/x(x−1) Ans.

Answer 2

The single logarithm expression for the given question is

= log$(x+1)^{2}$ ÷ log x(x-1)

Step-by-step explanation:

Given equation:

Logx - 2logx + 3log(x+1) - log($x^{2}$ -1)

Solution:

= -logx + 3log(x+1) - log($x^{2}$ -1)

= -logx + log($(x+1)^{3}$) - log($x^{2}$ -1)

= log$(x+1)^{3}$÷ logx - log($x^{2}$ -1)

= log$(x+1)^{3}$ ÷ log x($x^{2}$ -1)

=  log$(x+1)^{3}$ ÷ log x( x-1)(x+1)

= log$(x+1)^{2}$ ÷ log x(x-1)

Result:

The single logarithm expression for the given question is

= log$(x+1)^{2}$ ÷ log x(x-1)

(SPJ3)