Question-162814

Question Number 162814 by amin96 last updated on 01/Jan/22

Answered by MathsFan last updated on 01/Jan/22

l e t {l o g}_{3} x = u \to x = 3^{u}

2^{u} = \sqrt{3^{u}} + 1

2^{u} - \sqrt{3^{u}} = 4 - 3

2^{u} = 4 \to u = 2

x = 3^{2}

x = 9

Answered by Exauce last updated on 01/Jan/22

X =^{{l o g}_{3} 2} \sqrt{\sqrt{X} + 1}

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