Question: Solve: $3^{x-2} + 3^x = \dfrac{10}{9}$

Solution:
Given,

$3^{x-2} + 3^x = \dfrac{10}{9}$

$or, 3^x × 3^{-2} + 3^x = \dfrac{10}{9}$

$or, 3^x × \dfrac{1}{3²} + 3^x = \dfrac{10}{9}$

$or, 3^x \left ( \dfrac{1}{9} + 1 \right ) = \dfrac{10}{9}$

$or, 3^x \left ( \dfrac{10}{9} \right ) = \dfrac{10}{9}$

$or, 3^x eft (\dfrac{10 × 9}{9 × 10} \right ) = \dfrac{ 10 × 9}{9 × 10}$

$or, 3^x = 1$

$or, 3^x = 3^0$

$\therefore x = 0$
= Answer

Related Notes and Solutions:

Here is the website link to the notes of Indices.

#SciPiPupil