How do you solve $ln x + 2 ln 4 = ln 7 - ln x$ $ln x + 2ln4 = ln7 - ln x$ ?

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Answer 1 · 2016-02-02T11:37:03

$x = + \frac{\sqrt{7}}{4}$ $x=+sqrt7/4$

From the given:

$ln x + 2 ln 4 = ln 7 - ln x$ $ln x+2 ln 4=ln 7 - ln x$

$ln x + ln x = ln 7 - 2 ln 4$ $ln x+ln x=ln 7-2 ln 4$

$2 ln x = ln 7 - ln (4^{2})$ $2 ln x=ln 7-ln (4^2)$

${ln x}^{2} = ln (\frac{7}{16})$ $ln x^2=ln (7/16)$

$x^{2} = \frac{7}{16}$ $x^2=7/16$

$x = \pm \frac{\sqrt{7}}{4}$ $x=+-sqrt7/4$ take the positive value as answer

Have a nice day from the Philippines..

How do you solve ln x + 2ln4 = ln7 - ln xlnx+2ln4=ln7−lnxln x + 2ln4 = ln7 - ln x?