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.

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