Question

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

Answer 1

Answer:

log is 15 is the answer

hope this answers will be helpful to you kaira

Answer 2

$log(x + 1)^3- 2log (x + 1) + 3 log (x - 1)^2- log (x^3- 1)$ after expressing as a single logarithm is $log\frac{(x+1)(x-1)^6}{(x-1)(x^2+x+1)}$.

Given:

A logarithmic expression $log(x + 1)^3- 2log (x + 1) + 3 log (x - 1)^2- log (x^3- 1)$.

To Find:

The expression as a single logarithm.

Solution:

Write $-2log(x+1)$ as $-log(x+1)^2$ in the expression.

$log(x + 1)^3- log (x + 1)^2 + 3 log (x - 1)^2- log (x^3- 1)$

Now use the property $loga-logb=log\frac{a}{b}$ in the first two terms in the expression and simplify.

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

Use the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ in $log (x^3- 1)$.

$log(x + 1)+ 3 log (x - 1)^2- log (x-1)(x^2+x+1)$

Write $3 log (x - 1)^2$ as $log (x - 1)^6$ in the expression.

Use the property $loga+logb=logab$ and $loga-logb=\frac{loga}{logb}$ .

$log(x + 1)+ log (x - 1)^6- log (x-1)(x^2+x+1)=log(x+1)(x-1)^6- log (x-1)(x^2+x+1)\\=log\frac{(x+1)(x-1)^6}{(x-1)(x^2+x+1)}$

Thus, $log(x + 1)^3- 2log (x + 1) + 3 log (x - 1)^2- log (x^3- 1)$ after expressing as a single logarithm is $log\frac{(x+1)(x-1)^6}{(x-1)(x^2+x+1)}$.