Logarithm Question

• bonkysleuth

  5 Mar 08, 18:12

  Given that logb (xy2) =m and logb (x3y)=n, express logb  y/x and logb (square root xy) in terms of m and n. I can get the answer but I 'm not sure of how to present the working. could someone help? thanks.

• Shermanbok

  5 Mar 08, 20:08

  logb (xy^2) = m = logbx + 2logby..........(1)

  logb(x^3 y) = 3logbx + logby.....(2)

   

  3(1) - (2) = 3m-n = 5logby

  logby = (3m-n)/5

  From logby you can get logbx then you just sub into logb y/x = logby - logbx and

  to logb square root xy = 0.5logbx + o.5logby.

   

  Hope this helps!

• eagle

  5 Mar 08, 20:10

  (1) logb (xy2) =m
  (2) logb (x3y)=n
  
  
  (1)    becomes logb x + 2 logb y = m
  (2)    becomes 3logb x + logb y = n
  
  2 * (2) – (1) : 5 logb x = 2n – m
  (3) logb x = 0.2 (2n – m) = 0.4n – 0.2m
  Thus
  (4) logb y = n – 3 logb x = 0.6m – 0.2n
  
  logb  y/x = logb y - logb x = 0.8m – 0.6n
  logb sqrt(xy) = 0.5 (logb x + logb y) = 0.2m + 0.1n
  
  

• ^tamago^

  5 Mar 08, 20:31

  Given that log_b (xy²) = m and log_b (x³y)=n, express log_b (y/x) and log_b √(xy) in terms of m and n.
  
  log_b (xy²) = m
  ⇒ b^m = xy²
  
  log_b (x³y) = n
  ⇒ bⁿ = x³y
  
  Let α and β be the integer such that b^αm × b^βn = b^αm+βn = y/x
  ⇒ (xy²)^α × x³y^β = y/x
  ⇒ x^αy^2α × x^3βy^β = yx^-1
  ⇒ x^α+3βy^2α+β = yx^-1
  ⇒ α+3β = -1 ----- (1)
  ⇒ 2α+β = 1 ----- (2)
  sub (2) into (1): α+3β = -2α-β
  ⇒ soln: 3α = -4β
  Let α = 4 and β = -3
  ⇒ (xy²)� × (x³y)^-3 = y�/x�
  ⇒ b^4m × b^-3n = (y/x)�
  ⇒ y/x = b^4m/5 × b^-3n/5 = b^(4m-3n)/5
  ⇒ log_b (y/x) = (4m-3n)/5
  
  Let γ and δ be the integer such that b^γm × b^δn = b^γm+δn = √(xy)
  ⇒ (xy²)^γ × x³y^δ = y/x
  ⇒ x^γy^2γ × x^3δy^δ = (yx)^½
  ⇒ x^γ+3δy^2γ+δ = y^½x^½
  ⇒ γ+3δ = ½ ----- (1)
  ⇒ 2γ+δ = ½ ----- (2)
  sub (2) into (1): γ+3δ = 2γ+δ
  ⇒ soln: γ = 2δ
  Let γ = 2 and δ = 1
  ⇒ (xy²)² × x³y = (xy)�
  ⇒ (xy)� = b^2m+n
  ⇒ √(xy) = b^(2m+n)/10
  ⇒ log_b √(xy) = (2m+n)/10
• ^tamago^

  5 Mar 08, 20:34

  Mine is incredibly tedious compared to eagle's.