Question

(9/4)^x (8/27)^x-1=2/3 then...value of (x)

Answer 1

HELLO DEAR,

${ (\frac{9}{4}) }^{x} \times { \frac{8}{27} }^{x - 1} = \frac{2}{3} \\ \\ = > {( \frac{2}{3} )}^{ - 2x} \times {( \frac{2}{3} )}^{3(x - 1)} = \frac{2}{3} \\ \\ = { (\frac{2}{3} )}^{ - 2x + 3x - 3} = \frac{2}{3} \\ \\ = > x - 3 = 1 \\ \\ = > x = 4$


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

The value of x in the expression $(\frac{9}{4})^x \times (\frac{8}{27})^{x-1}=\frac{2}{3}$ is 4.    

Step-by-step explanation:

Consider the given expression

$(\frac{9}{4})^x \times (\frac{8}{27})^{x-1}=\frac{2}{3}$

$(\frac{3^2}{2^2})^x \times (\frac{2^3}{3^3})^{x-1}=\frac{2}{3}$

Using property of exponents, $(a^n)^m=a^{mn}$ , we get,

$(\frac{3}{2})^2x \times (\frac{2}{3})^{3(x-1)}=\frac{2}{3}$

Using property of exponents, $a^m=(\frac{1}{a})^{-m}$

$(\frac{2}{3})^{-2x} \times (\frac{2}{3})^{3(x-1)}=\frac{2}{3}$

Using property of exponents, $a^n \times a^m=a^{n+m}$

$(\frac{2}{3})^{-2x+3(x-1)}=\frac{2}{3}$

Now comparing power in both sides, $a^m=a^n \Rightarrow m=n$

${-2x+3(x-1)}=1$

Solving for x, we get,

$-2x+3x-3=1$

$x=4$

Thus, the value of x in the expression $(\frac{9}{4})^x \times (\frac{8}{27})^{x-1}=\frac{2}{3}$ is 4.