If x, y are real numbers such that 3^{(x/y)+1 | vedclass

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

Question

(D) Given the equation: $3^{(x/y)+1} - 3^{(x/y)-1} = 24$Let $u = x/y$. Then the equation becomes $3^{u+1} - 3^{u-1} = 24$.Factor out $3^{u-1}$:$3^{u-1

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Given the equation: $3^{(x/y)+1} - 3^{(x/y)-1} = 24$Let $u = x/y$. Then the equation becomes $3^{u+1} - 3^{u-1} = 24$.Factor out $3^{u-1}$:$3^{u-1} \cdot (3^2 - 1) = 24$$3^{u-1} \cdot (9 - 1) = 24$$3^{u-1} \cdot 8 = 24$$3^{u-1} = 3$Since $3^{u-1} = 3^1$,we have $u - 1 = 1$,which implies $u = 2$.Thus,$x/y = 2$.We need to find the value of $\frac{x+y}{x-y}$. Dividing the numerator and denominator by $y$:$\frac{x/y + 1}{x/y - 1} = \frac{2 + 1}{2 - 1} = \frac{3}{1} = 3$.

If $x, y$ are real numbers such that $3^{(x/y)+1} - 3^{(x/y)-1} = 24$,then the value of $(x+y)/(x-y)$ is

Similar Questions

If ${2^x} = {4^y} = {8^z}$ and $xyz = 288$,then find the value of $\frac{1}{{2x}} + \frac{1}{{4y}} + \frac{1}{{8z}}$.

If $\frac{\log x}{b - c} = \frac{\log y}{c - a} = \frac{\log z}{a - b}$,then which of the following is true?

$0.5737373...... = $

The square root of $\frac{(0.75)^3}{1-0.75}+[0.75+(0.75)^2+1]$ is

The value of $a^{\log_b x}$,where $a = 0.2$,$b = \sqrt{5}$,and $x = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ to $\infty$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module