Simultaneous Equations (Stuck on the algebra)

Question

Question: Solve the following simultaneous equations for real values of x and y $$ \left\{ \begin{array}{l} 9^{2x+y} - 9^x \times 3^y = 6 \\ \log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3 \end{array} \right. $$

What I have attempted; for the first equation $$ 9^{2x+y} - 9^x \times 3^y = 6 \\ 9^{2x+y} - 3^{2x} \times 3^y = 6 \\ 9^{2x+y} - 3^{2x+y} - 6 = 0 $$ Let $z = 3^{2x+y}$. Then $$ z^2 - z - 6 = 0 \iff (z-3)(z+2) = 0 \iff z = 3, z \ne -2 $$ where $$ 3^{2x+y} = 3 \iff y = 1-2x. $$ Now using the second equation I get $$ \log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3 $$ Substituting $y = 1-2x$ $$ \log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3. $$ Now this is the part I am stuck on , how do I solve for $x$ algebraically?

Answer 1

Observe $$\log_{x+1}(y+3)+\log_{x+1}(y+x+4)=\log_{x+1}[(y+3)(y+x+4)]\text{.}$$ Now we have $$\log_{x+1}[(y+3)(y+x+4)] = 3$$ Take both as a power of $x+1$: $$(x+1)^{\log_{x+1}[(y+3)(y+x+4)] } = (x+1)^{3}$$ giving $$(y+3)(y+x+4)=(x+1)^{3}\text{.}$$ [since $b^{\log_{b}(x)} = x$]

Answer 2

we have $$(3^{2x+y})^2-3^{2x+y}=6$$ and $$log_{x+1}(y+3)(y+x+4)=3$$ the last equation can be written as $$(y+3)(y+x+4)=(x+1)^3$$ from the quadratic equation we get $$2x+y=1$$ can you proceed?

Answer 3

Writing the last equation

$$ \log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3 $$

as

$$ \log_{x+1}[(4-2x)(5-x)] = 3 $$

you can get both term as exponents of $x+1$, so that

$$ (4-2x)(5-x) = (x+1)^3 $$

This equation can be easily solved, leading to $x=1$.

Answer 4

You finished with $$\log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3$$ Go to natural logarithms (the only ones I know !); this will then write $$\frac{\log (4-2 x)}{\log (x+1)}+\frac{\log (5-x)}{\log (x+1)}=3$$ Multiply by the denominator appearing in the lhs (hoping that $x\neq -1$) and simplify.

So, $$\log (4-2 x)+\log (5-x)=3\log(x+1)$$ that is to say $$(4-2x)\times (5-x)=(x+1)^3$$ Expand and simplify to get $$x^3+x^2+17 x-19=0$$ where $x=1$ is an obvious solution. Perform the long division to get $$x^3+x^2+17 x-19=(x-1)(x^2+2 x+19)=0$$ The quadratic term does not show real solutions (in the complex domain, they would be $(x_{1,2}=-1\pm3 i \sqrt{2})$.

So $x=1$ is your solution and you were about to get it.

Happy New Year !!!!