# Determine the number of solutions for a system of equation

#### anemone

##### MHB POTW Director
Staff member
Determine the number of solutions of the simultaneous equations $x^2+y^3=29$ and $\log_3x\log_2y=1$.

#### Amer

##### Active member
Determine the number of solutions of the simultaneous equations $x^2+y^3=29$ and $\log_3x\log_2y=1$.
$\log_3 x = a \Rightarrow x = 3^a$
$\log_2 y = b \Rightarrow y = 2^b$

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

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

$f(x) = 3^{2/x} + 2^{3x} - 29$
$f(1) = 9 + 8 - 29 < 0$
$f(2) = 3 + 64 - 29 > 0$ 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

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#### anemone

##### MHB POTW Director
Staff member
$\log_3 x = a \Rightarrow x = 3^a$
$\log_2 y = b \Rightarrow y = 2^b$

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

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

$f(x) = 3^{2/x} + 2^{3x} - 29$
$f(1) = 9 + 8 - 29 < 0$
$f(2) = 3 + 64 - 29 > 0$ 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!