Express as a sum and/or difference of logarithms: ln x 2 - 4 x - 5 x 2 + 6 x + 8

Given, ln x 2 - 4 x - 5 x 2 + 6 x + 8 = ln x 2 - 5 x + x - 5 x 2 + 4 x + 2 x + 8 = ln x x - 5 + 1 x - 5 x x + 4 + 2 x + 4 = ln x + 1 x - 5 x + 2 x + 4 Now, applying the property of logarithm of a quotient, we have ln x 2 - 4 x - 5 x 2 + 6 x + 8 = ln x + 1 x - 5 - ln x + 2 x + 4 ∵ ln p q = ln p - ln q , where p , q > 0 = ln x + 1 + ln x - 5 - ln x + 2 + ln x + 4 ∵ ln p q = ln p + ln q , where p , q > 0 Hence, ln x 2 - 4 x - 5 x 2 + 6 x + 8 = ln ( x - 5 ) + ln ( x + 1 ) - ln ( x + 4 ) - ln ( x + 2 )

Express as a sum and/or difference of logarithms: ln x 2 - 10 x - 24 x 3 - 4 x

Given, ln x 2 - 10 x - 24 x 3 - 4 x = ln x 2 - 12 x + 2 x - 24 x x 2 - 4 = ln x x - 12 + 2 x - 12 x x - 2 x + 2 = ln x + 2 x - 12 x x - 2 x + 2 = ln x - 12 x x - 2 Now, applying the property of logarithm of a quotient, we have ln x 2 - 10 x - 24 x 3 - 4 x = ln x - 12 - ln x x - 2 ∵ ln p q = ln p - ln q , where p , q > 0 = ln x - 12 - ln x + ln x - 2 ∵ ln p q = ln p + ln q , where p , q > 0 Hence, ln x 2 - 10 x - 24 x 3 - 4 x = ln x - 12 - ln x - ln x - 2

Solve the given exponential equation: 3 2 x + 1 + 3 = 10 × 3 x

Given equation is 3 2 x + 1 + 3 = 10 × 3 x ⇒ 3 2 x · 3 + 3 = 10 × 3 x ⇒ 3 2 x × 3 - 10 × 3 x + 3 = 0 ⇒ 3 x 3 x + 1 - 10 + 3 = 0 3 x + 1 = 7 [Since 3 x = 0 is not possible] Now, taking logarithm on both sides of the equation, we have x + 1 log 3 = log 7 ∵ log p q = q log p , where p > 0 and q ∈ R ⇒ x = log 7 log 3 - 1 ⇒ x = 0 . 845098040 0 . 477121255 - 1 = 0 . 7712 (approximately) Hence, x = 0 . 7712 (approximately).