If log a N log c N = log a N − log b N log b N − log c N where N > 0 and N ≠ 1 , a , b , c > 0 and not equal to 1 , then prove that b 2 = ac .

Given that log a N log c N = log a N − log b N log b N − log c N where N > 0 and N ≠ 1 , a , b , c > 0 and not equal to 1 . Apply base change ratio property of logarithm , we have log N log a log N log c = log N log a - log N log b log N log b - log N log c ⇒ log c log a = log b - log a log a log b × log b log c log c - log b ⇒ log N c log N a = log N b - log N a log N c - log N b log N c log N a ⇒ log N c - log N b = log N b - log N a ⇒ 2 log N b = log N a + log N c ⇒ log N b 2 = log N a c ⇒ b 2 = a c

Solve for x : log 5 120 + ( x - 3 ) - 2 · log 5 1 - 5 x - 3 = - log 5 0 . 2 - 5 x - 4

Given that log 5 120 + ( x - 3 ) - 2 · log 5 1 - 5 x - 3 = - log 5 0 . 2 - 5 x - 4 ⇒ log 5 120 + x - 3 - log 1 - 5 x - 3 2 = - log 5 0 . 2 1 - 5 x - 3 ⇒ log 5 120 + x - 3 - 2 log 1 - 5 x - 3 = - log 5 0 . 2 - log 1 - 5 x - 3 = - log 5 1 - 5 x - 3 + log 5 5 ⇒ log 5 120 - log 5 5 + x - 3 = log 5 1 - 5 x - 3 ⇒ log 5 24 + ( x - 3 ) = log 5 1 - 5 x - 3 ⇒ log 5 1 - 5 x - 3 24 = x - 3 ⇒ 1 - 5 x - 3 24 = 5 x - 3 ⇒ 25 5 x - 3 = 1 ⇒ 5 x - 3 = 1 5 2 ⇒ x - 3 = - 2 ⇒ x = 1

Solve for x : log 4 + 1 + 1 2 x log 3 = log ( 3 x + 27 ) .

Given that log 4 + 1 + 1 2 x log 3 = log ( 3 x + 27 ) ⇒ log 4 + log 3 + log 3 1 2 x = log 3 1 x + 27 ⇒ log 4 × 3 × 3 1 / 2 x = log 3 1 / x + 27 ⇒ 12 · 3 1 2 x = 3 1 x + 27 Let 3 1 2 x = y ⇒ y 2 - 12 y + 27 = 0 ⇒ y 2 - 9 y - 3 y + 27 = 0 ⇒ y - 3 y - 9 = 0 ⇒ y = 9 or y = 3 ⇒ x = 1 4 and x = 1 2 But 3 x is only defined for x ∈ N ≥ 2 , it has no solution. Hence, x ∈ ϕ

Find x satisfying the equation log 2 1 + 4 x + log 2 1 - 4 x + 4 = 2 log 2 2 x - 1 - 1 .

Given that log 2 1 + 4 x + log 2 1 - 4 x + 4 = 2 log 2 2 x - 1 - 1 ⇒ log 2 x + 4 x + log 2 x x + 4 = 2 log 2 3 - x x - 1 ⇒ 2 log 2 x + 4 x = 2 log 2 3 - x x - 1 ⇒ log x + 4 x = ± log 3 - x x - 1 ⇒ x + 4 x = 3 - x x - 1 ⇒ x = ± 2 ⇒ x = 2 only solution and satisfying domain . also x + 4 x = x - 1 3 - x ⇒ x = ± 6 ⇒ x = 6 is the solution and satisfying the domain .

Solve for x : log 2 ( 4 - x ) + log ( 4 - x ) · log x + 1 2 - 2 log 2 x + 1 2 = 0 .

Given that log 2 ( 4 - x ) + log ( 4 - x ) log x + 1 2 - 2 log 2 x + 1 2 = 0 Let log ( 4 - x ) = A and log x + 1 2 = B ⇒ A 2 + A B - 2 B 2 = 0 ⇒ ( A + 2 B ) ( A - B ) = 0 ⇒ A = - 2 B or A = B ⇒ log ( 4 - x ) = - 2 log x + 1 2 or log ( 4 - x ) = log x + 1 2 ⇒ 4 - x = 1 x + 1 2 2 or ( 4 - x ) = x + 1 2 ⇒ x 2 + x + 1 4 4 - x = 1 or x = 7 4 ⇒ 4 x 2 - x 3 + 4 x - x 2 + 1 - x 4 = 1 ⇒ 3 x 2 - x 3 + 15 4 x = 0 ⇒ x x 2 - 3 x - 15 4 = 0 ⇒ x = 0 or 4 x 2 - 12 x - 15 = 0 ⇒ x = 0 or x = 12 ± 144 + 240 8 = 12 ± 384 8 = 12 ± 16 × 24 8 ⇒ x = 0 , 7 4 , 3 + 24 2