Let (x0, y0) be the solution of the following equations (2x)ln2 = (3y)ln3 3ln x = 2ln y Then x0 is

Engineering

Mathematics

Properties of Logarithm

Question

Let (x₀, y₀) be the solution of the following equations

(2x)^ln2 = (3y)^ln3

3^{ln x} = 2^{ln y}

Then x₀ is

$\frac{1}{2}$

$\frac{1}{6}$

$\frac{1}{3}$

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(2x)^{ln 2} = (3y)^{ln 3}

ln 2 (ln 2 + ln x) = ln 3 (ln 3 + ln y)

ln 2 . ln x – ln 3 ln y = (ln 3)² – (ln 2)² .....(1)

3^{ln x} = 2^{ln y}

ln x . ln 3 = lny . ln 2

$ℓn y = ℓn x \frac{ℓn 3}{ℓn 2}$ .....(2)

Solving (1) & (2)

ln x = – ln 2 ⇒ x = $\frac{1}{2}$

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