Question-159507

Question Number 159507 by cortano last updated on 18/Nov/21

Commented by tounghoungko last updated on 18/Nov/21

\frac{5 . 5^{- 2 x} + 5^{3} . 5^{- 2 x}}{\frac{5}{8} . 2^{2 x}} = 4

\Leftrightarrow 5^{- 2 x} + 5^{2} . 5^{- 2 x} = \frac{1}{2} . 2^{2 x}

\Leftrightarrow 26 . 5^{- 2 x} = \frac{1}{2} . 2^{2 x}

\Leftrightarrow 52 = 10^{2 x} \Rightarrow x = \frac{1}{2} \log_{10} (52)

Answered by Tokugami last updated on 18/Nov/21

\frac{5^{1 - 2 x} + \sqrt{25^{3 - 2 x}}}{2^{2 x - 1} + 2^{2 x - 3}} = 4

\frac{5^{1 - 2 x} + \sqrt{25^{3 - 2 x}}}{2^{2 x - 1} + 2^{2 x - 3}} = 2^{2}

5^{1 - 2 x} + \sqrt{25^{3 - 2 x}} = 2^{2 x - 1 + 2} + 2^{2 x - 3 + 2}

5^{1 - 2 x} + 5^{3 - 2 x} = 2^{2 x + 1} + 2^{2 x - 1}

5^{- 2 x} (5^{1} + 5^{3}) = 2^{2 x} (2^{1} + 2^{- 1})

\frac{130}{5^{2 x}} = 2^{2 x} (\frac{5}{2})

\frac{260}{5} = 10^{2 x}

52 = 10^{2 x}

\ln (52) = 2 x \ln (10)

2 x = \frac{\ln 52}{\ln 10} = \log_{10} (52)

x = \frac{1}{2} \log_{10} (52)

x \approx 0.858

Answered by qaz last updated on 18/Nov/21

\frac{5^{1 - 2 x} \sqrt{25^{3 - 2 x}}}{2^{2 x - 1} + 2^{2 x - 3}} = \frac{5^{1 - 2 x} + 5^{3 - 2 x}}{2^{2 x - 1} + 2^{2 x - 3}} = \frac{(5 + 5^{3}) 5^{- 2 x}}{(2^{- 1} + 2^{- 3}) 2^{2 x}} = 208 \cdot 10^{- 2 x} = 4

\Rightarrow x = \frac{1}{2} \lg 52

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