Finding number of solutions of 2 simultaneous equations.

Question

There are 2 equations
$x^2+y^3=29$ and $\log_3x.\log_2y=1$.
Find the number of solutions in $x$ and $y$ where $x,y \in \Bbb R$

Answer 1

Rewrite the second equation as $$ \ln x\cdot \ln y=\ln 2\cdot \ln 3.$$ To even speak about logs, we need $x,y$ to be positive. Also, $\ln2\ln3$ is positive hence either both $x,y$ are $<1$ r both are $>1$. The first option makes $x^2+y^3<2$ and is rejected. We conclude $1<x<\sqrt{28}$. We can eliminate $y$ using the first equation to arrive at $$ \ln x\cdot\ln\sqrt[3]{29-x^2}=\ln 2\ln3$$ or $$ \ln x^2\cdot \ln(29-x^2)=6\ln2\ln3.$$ With $z:=x^2-\frac{29}2$ (ranging from $-\frac{27}2$ to $\frac{27}2$ as $x$ ranges from $1$ to $\sqrt{28}$), this equation is of the form $$\tag1f(z)f(-z)=c $$ where $f(z)=\ln(z+\frac{29}2)$ is a strictly increasing function. The solutions of $(1)$ are symmetric around $0$. Numerically, we verify that for $z=0$, the left hand side is greater than the right hand side. On the other hand, $f(-\frac{27}2)=0$ so that we conclude from the intermediate value theorem that there is at least one solution with $z\in(0,\frac{27}2)$. Assume there are two positive solutions of $(1)$. Then by Rolle, there must be an intermediate (hence also positive) point where the derivative of the left hand side is $=0$, i.e., for some $z\in(0,\frac{27}2)$, $$ f'(z)f(-z)=f(z)f'(-z)$$ i.e., $$ \frac{\ln(-z+\frac{29}2)}{z+\frac{29}2}=\frac{\ln(z+\frac{29}2)}{-z+\frac{29}2}$$ or $ g(-z)=g(z)$ where $g(z)=(z+\frac{29}2)\ln(z+\frac{29}2)$ . In the range in question, $g$ is strictly increasing (product of positive increasing functions), hence $g(-z)=g(z)$ has only one solution, and that is the trivial solution $z=0$.

We conclude that $(1)$ and thereby the original equation system has exactly

two

solutions in the range considered.

Answer 2

Here is a solution from dan fulea.

This is an ugly problem, only a numerical solution is possible.
After setting $x=3u, y=21/u$ we have to solve $f(u):=32u+23/u−29=0$,
$f$ is convex, $f(1/2)=38>0$, $f(1)=−12<0$, $f(3/2)=2>0$, so we expect a solution between $1/2$ and $1$, and an other one between $1$ and $3/2$,
$pari/gp$ delivers for solve
$( u=0.1,1, 3^{(2*u)}+2^{(3/u)} - 29 )$ the result $0.64718824239603875658029040340778233317$
and for other solve
$( u=1,1.5, 3^{(2*u)}+2^{(3/u)} - 29 )$ the result $1.4623174037485863183265033005868559095$ ...