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Solve \(\log (x - 4) - \log (3x - 10) = \log \frac{1}{x}\)

 Jun 5, 2020
 #1
avatar+26600 
+1

log (x-4)/log (3x-10) = log 1/x

 

(x-4 ) = (3x-10)/x

x^2 --7x+10 = 0

(x-5)(x-2) = 0      x = 5    or 2 (throw out)

 Jun 5, 2020
 #2
avatar+9998 
+1

Solve \( \)\( \)\(\log (x - 4) - \log (3x - 10) = \log \frac{1}{x} \)

 

Hello Guest!

 

 \(\color{BrickRed}\log (x - 4) - \log (3x - 10) = \log \frac{1}{x}\\ \log \large \frac{x-4}{3x-10}=log \frac{1}{x}\\ \large \frac{x-4}{3x-10}= \frac{1}{x}\\ x(x-4)=3x-10\)

\(x^2-4x-3x+10=0\\ x^2-7x+10=0\\ x=\frac{7}{2}\pm\sqrt{(\frac{7}{2})^2-10}\\ x=\frac{7}{2}\pm\sqrt{2.25}\\ x=\frac{7}{2}\pm\frac{3}{2}\)

\(\color{blue}x_1=5\\ x_2=2\ (throw\ out)\)

laugh  !

 Jun 5, 2020
edited by asinus  Jun 5, 2020
edited by asinus  Jun 5, 2020

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