Logarithm Question

bonkysleuth SG

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 SG

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 SG

(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^ SG

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)⁴ × (x³y)^-3 = y⁵/x⁵
⇒ b^4m × b^-3n = (y/x)⁵
⇒ 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)⁵
⇒ (xy)⁵ = b^2m+n
⇒ √(xy) = b^(2m+n)/10
⇒ log_b √(xy) = (2m+n)/10

^tamago^ SG

Mine is incredibly tedious compared to eagle's.