Number of solutions to x² + y³ = 29 and log₃x log₂y = 1

[anemone]

Determine the number of solutions of the simultaneous equations $x^2+y^3=29$ and $\log_3x\log_2y=1$.

 

[Amer]

Determine the number of solutions of the simultaneous equations $x^2+y^3=29$ and $\log_3x\log_2y=1$.

Spoiler

$\log_3 x = a \Rightarrow x = 3^a $
$\log_2 y = b \Rightarrow y = 2^b $

$a.b =1 $
$(3^a)^2 + (2^b)^3 = 29 $

$(3^{2/b}) + (2^{3b}) = 29 $

$f(x) = 3^{2/x} + 2^{3x} - 29 $
$f(1) = 9 + 8 - 29 < 0 $
$f(2) = 3 + 64 - 29 > 0 $ we have a zero at (1,2) interval
$f(1/2) = 3^4 + 2^{3/2} - 29 > 0 $ we have anther zero at (1/2 , 1 )

numbers larger than 2 , $2^{3x}$ is more than 29 so there is not any zero (2, infinity)
for x in (0,1/2) $3^{2/x}$ is more than 29 f(x) is positive
for negative numbers $3^{2/x} + 2^{3x} $ less than 29
so I think we have just two zeros
two solutions

 

[anemone]

Spoiler

$\log_3 x = a \Rightarrow x = 3^a $
$\log_2 y = b \Rightarrow y = 2^b $

$a.b =1 $
$(3^a)^2 + (2^b)^3 = 29 $

$(3^{2/b}) + (2^{3b}) = 29 $

$f(x) = 3^{2/x} + 2^{3x} - 29 $
$f(1) = 9 + 8 - 29 < 0 $
$f(2) = 3 + 64 - 29 > 0 $ we have a zero at (1,2) interval
$f(1/2) = 3^4 + 2^{3/2} - 29 > 0 $ we have anther zero at (1/2 , 1 )

numbers larger than 2 , $2^{3x}$ is more than 29 so there is not any zero (2, infinity)
for x in (0,1/2) $3^{2/x}$ is more than 29 f(x) is positive
for negative numbers $3^{2/x} + 2^{3x} $ less than 29
so I think we have just two zeros
two solutions

Hi Amer, thanks for participating and yes, your answer is correct and your method by using the substitution skill seems awesome too!