The number of solution pairs (x, y) of the si | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given equations are:$1) \log _{1 / 3}(x+y)+\log _3(x-y)=2$$2) 2^{y^2}=512^{x+1}$From equation $(1)$:$-\log _3(x+y)+\log _3(x-y)=2$$\log _3\left(\f

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given equations are:$1) \log _{1 / 3}(x+y)+\log _3(x-y)=2$$2) 2^{y^2}=512^{x+1}$From equation $(1)$:$-\log _3(x+y)+\log _3(x-y)=2$$\log _3\left(\frac{x-y}{x+y}\right)=2$$\frac{x-y}{x+y}=3^2=9$$x-y=9x+9y$$-8x=10y \Rightarrow y = -\frac{4}{5}x$From equation $(2)$:$2^{y^2}=(2^9)^{x+1} = 2^{9(x+1)}$$y^2=9(x+1)$Substitute $y = -\frac{4}{5}x$ into $y^2=9(x+1)$:$\left(-\frac{4}{5}x\right)^2=9(x+1)$$\frac{16}{25}x^2=9x+9$$16x^2=225x+225$$16x^2-225x-225=0$$(16x+15)(x-15)=0$So,$x=15$ or $x=-\frac{15}{16}$.For $x=15$,$y=-\frac{4}{5}(15)=-12$. Check domain: $x+y=15-12=3 > 0$ and $x-y=15-(-12)=27 > 0$. This is a valid solution.For $x=-\frac{15}{16}$,$y=-\frac{4}{5}(-\frac{15}{16})=\frac{3}{4}$. Check domain: $x+y=-\frac{15}{16}+\frac{12}{16}=-\frac{3}{16} Thus,there is only $1$ solution pair $(15, -12)$.

The number of solution pairs $(x, y)$ of the simultaneous equations $\log _{1 / 3}(x+y)+\log _3(x-y)=2$ and $2^{y^2}=512^{x+1}$ is

Similar Questions

Find the solution of the equation $9^x - 2^{x + 1/2} = 2^{x + 3/2} - 3^{2x - 1}$.

If $x = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$,then find the value of $3x^2 + 4xy - 3y^2$.

The sum of squares of all the real solutions of the equation $\log_{(x+1)}(2x^2 + 5x + 3) = 4 - \log_{(2x+3)}(x^2 + 2x + 1)$ is equal to . . . . . . .

Evaluate: $\frac{{[4 + \sqrt{15}]}^{3/2} + {[4 - \sqrt{15}]}^{3/2}}{{[6 + \sqrt{35}]}^{3/2} - {[6 - \sqrt{35}]}^{3/2}}$

Let $p = 99$ and $q = 101$. Define $p_1 = \log_{10} \left(\frac{p+q}{2}\right)$ and $q_1 = \frac{1}{2}(\log_{10} p + \log_{10} q)$,and $p_2 = \log_{10} \left(\frac{p_1+q_1}{2}\right)$,$q_2 = \frac{1}{2}(\log_{10} p_1 + \log_{10} q_1)$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module