3-logx-2-logx-1-2-logx-1-2-3-logx-1-x-

Question Number 10492 by ABD last updated on 14/Feb/17

3^{l o g x} - 2^{l o g x - 1} = 2^{l o g x + 1} - 2 \times 3^{l o g x - 1} \Rightarrow x = ?

Answered by mrW1 last updated on 14/Feb/17

3^{l o g x} - \frac{2^{\log x}}{2} = 2 \times 2^{l o g x} - 2 \times \frac{3^{\log x}}{3}

\frac{5}{3} \times 3^{l o g x} = \frac{5}{2} \times 2^{l o g x}

3^{l o g x - 1} = 2^{l o g x - 1}

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

\log x - 1 = 0

\log x = 1

x = 10

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