Is it a problem

Algebra Level 3

Let ( x , y ) (x , y) be the solution of the following system of equations:

{ ( 2 x ) ln ⁡ 2 = ( 3 y ) ln ⁡ 3 3 ln ⁡ x = 2 ln ⁡ y \begin{cases} (2x)^{\ln 2} = (3y)^{\ln 3} \\ 3^{\ln x} = 2^{\ln y } \\ \end{cases}

What is x x ?

1/3 1/6 1/2 6

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Sushil Kumar by Sushil Kumar , 29, India

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2 solutions

Chew-Seong Cheong
Jul 29, 2015

{ ( 2 x ) ln ⁡ 2 = ( 3 y ) ln ⁡ 3 . . . ( 1 ) 3 ln ⁡ x = 2 ln ⁡ y ⇒ x ln ⁡ 3 = y ln ⁡ 2 . . . ( 2 ) \begin{cases} (2x)^{\ln{2}} = (3y)^{\ln{3}} & & ...(1) \\ 3^{\ln{x}} = 2^{\ln{y}} & \Rightarrow x^{\ln{3}} = y^{\ln{2}} & ...(2) \end{cases}

( 1 ) : 2 ln ⁡ 2 x ln ⁡ 2 = 2 ln ⁡ 3 ln ⁡ 2 ˙ ln ⁡ 3 y ln ⁡ 2 ln ⁡ 3 ln ⁡ 2 = 2 ( ln ⁡ 3 ) 2 ln ⁡ 2 x ( ln ⁡ 3 ) 2 ln ⁡ 2 ⇒ x ln ⁡ 2 − ( ln ⁡ 3 ) 2 ln ⁡ 2 = 2 ( ln ⁡ 3 ) 2 ln ⁡ 2 − ln ⁡ 2 = ( 1 2 ) ln ⁡ 2 − ( ln ⁡ 3 ) 2 ln ⁡ 2 ⇒ x = 1 2 \begin{aligned} (1): \quad 2^{\ln{2}} x^{\ln{2}} & = 2^{\frac{\ln{3}}{\ln{2}}\dot{} \ln{3}} y^{\frac{\ln{2}\ln{3}}{\ln{2}}} \\ & = 2^{\frac{(\ln{3})^2}{\ln{2}}} x^{\frac{(\ln{3})^2}{\ln{2}}} \\ \Rightarrow x^{\ln{2} - \frac{(\ln{3})^2}{\ln{2}}} & = 2^{\frac{(\ln{3})^2}{\ln{2}}-\ln{2}} \\ & = \left(\frac{1}{2}\right)^{\ln{2} - \frac{( \ln{3})^2}{\ln{2}}} \\ \Rightarrow x & = \boxed{\frac{1}{2}} \end{aligned}

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Moderator note:

Hm, I don't quite see how (2) only yields those 2 cases to consider. Doesn't x = 1 , y = 1 x = 1, y = 1 also satisfy the condition?

Sorry, I was too eager to fix a solution. My solution revised.

Chew-Seong Cheong - 5 years, 10 months ago

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Yes, this looks much better.

Note that you also need to show that the exponent ln ⁡ 2 − ln ⁡ 2 3 ln ⁡ 2 \ln 2 - \frac{ \ln ^2 3 } { \ln 2 } is non-zero, in order to conclude that x = 1 2 x = \frac{1}{2} .

Calvin Lin Staff - 5 years, 10 months ago
Sushil Kumar
Jul 29, 2015

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

log(2x)^ln 2 = log(3y)^ln3

ln 2 . log(2x) = ln 3. log(3y)

this equation will get satisfied when x and y will be zero

therefore

log (2x)=0

implies that

2x = 1 (taking antilog )

therefore x=1/2

similarly y = 1/3

now it's time to solve other equation

3^ln x = 2^ln y

taking log on both side

log 3^ln x = log 2^ln y

ln x. log 3 = ln y .log 2

putting value of x and y

log 1/2 . log 3 = ln 1/3 . log 2

ln 2^-1 . log 3 = ln 3^-1 . log 2

-ln 2 . log 3 = -ln 3. log 2

(because log a ^n = n log a ) both side are equal the value of x

there fore x = 1/2

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I do not understand "this equation will get satisfied when x and y will be zero"

Are you saying that is the only possible solution? If so, why?

Calvin Lin Staff - 5 years, 10 months ago

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