Use logarithmic rules to evalue the expression. `ln((6sqrtx)/(x^2sqrt(1-x^2)))`Use logarithmic rules to evalue the expression. Rewrite the following expression in terms of Inx, In(1-x) and In(1 +...

Use logarithmic rules to evalue the expression. `ln((6sqrtx)/(x^2sqrt(1-x^2)))`

Use logarithmic rules to evalue the expression. Rewrite the following expression in terms of Inx, In(1-x) and In(1 + x)

`ln((6sqrtx)/(x^2sqrt(1-x^2)))`

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You need to use the following logarithmic identity, converting the logarithm of quotient into difference of logarithms, such that:

`ln(a/b) = ln a - ln b`

Reasoning by analogy yields:

`ln((6sqrt x)/(x^2sqrt(1-x^2))) = ln (6sqrt x) - ln(x^2sqrt(1 - x^2))`

Converting the logarithm of product into the sum of logarithms yields:

`ln(a*b) = ln a + ln b`

Reasoning by analogy yields:

`ln (6sqrt x) = ln 6 + ln sqrt x`

`ln(x^2sqrt(1 - x^2)) = ln x^2 + ln(sqrt(x^2 - 1))`

Converting the square root into a power yields:

`ln sqrt x = ln (x^(1/2))`

You need to use the following logarithmic identity, such that:

`ln(a^b) = b*ln a`

Reasoning by analogy yields:

`ln (x^(1/2)) = (1/2) ln x`

`ln x^2 = 2 ln x`

`ln(sqrt(x^2 - 1)) = (1/2)ln(x^2 - 1)`

You need to convert the difference of squares `x^2 - 1` into a product, such that:

`x^2 - 1 = (x - 1)(x + 1)`

`ln(x^2 - 1) = ln (x - 1)(x + 1) => ln(x^2 - 1) = ln(x - 1) + ln(x+1)`

`ln((6sqrt x)/(x^2sqrt(1-x^2))) = ln 6 +(1/2) ln x - 2 ln x - (1/2)ln(x - 1) - (1/2)ln(x + 1)`

`ln((6sqrt x)/(x^2sqrt(1-x^2))) = ln 6 - (3/2)ln x - (1/2)ln(x - 1) - (1/2)ln(x + 1)`

Hence, evaluating the given expression, using logarithmic identities, yields `ln((6sqrt x)/(x^2sqrt(1-x^2))) = ln 6 - (3/2)ln x - (1/2)ln(x - 1) - (1/2)ln(x + 1).`

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