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horsena [70]
3 years ago
15

Solve for x logx+log(x-4)=2log5

Mathematics
2 answers:
sleet_krkn [62]3 years ago
8 0
\log{x} + \log{(x-4)} = 2 \log{5}\\\\
\log{\big(x(x-4)\big)} = \log{5^2}\\\\
\log{(x^2-4x)} = \log{25} \\\\
x^2-4x = 25\\\\
x^2-4x-25=0 \\\\
x = \frac{4 \pm\sqrt{16+100}}{2} \\\\
x = \frac{4\pm2\sqrt{29}}{2} \\\\
x = 2 \pm \sqrt{29} \\\\
\text{But } x > 0 \implies x = 2 + \sqrt{29}
allochka39001 [22]3 years ago
8 0
D:x>0\wedge x>4\\D:x>4\\\\
\log x+\log(x-4)=2\log5\\
\log x(x-4)=\log25\\
x(x-4)=25\\
x^2-4x-25=0\\
x^2-4x+4-29=0\\
(x-2)^2=29\\
x-2=\sqrt{29} \vee x-2=-\sqrt{29}\\
x=2+\sqrt{29} \vee x=2-\sqrt{29}\\
2-\sqrt{29}\not>4\\
\Downarrow\\
\boxed{x=2+\sqrt{49}}
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63/2 or 31.5

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Andrews [41]

Hello and Good Morning/Afternoon:

<u>Let's take this problem step-by-step:</u>

<u>First off, let's write the line in point-slope form:</u>

  \rm \hookrightarrow (y-y_0)=m(x-x_0)

  • (x₀, y₀) any random point on the line
  • 'm' is the value of the slope

<u>Let's calculate the slope:</u>

 \rm \hookrightarrow Slope = \frac{y_2-y_1}{x_2-x_1}

  • (x₁, y₁): any random point on the line                                     ⇒   (-2, -6)
  • (x₂,y₂): any random point on the line that is not (x₁, y₁)         ⇒   (2, -3)

                 \rm \hookrightarrow slope = \frac{-3--6}{2--2} =\frac{3}{4}

<u>Now that we found the slope, let's put it into the point-slope form</u>

  ⇒ we need (x₀, y₀) ⇒ let's use (2,-3)

     (y-(-3))=\frac{3}{4} (x-2)\\y+3=\frac{3}{4} (x-2)

<u>The equation, however, could also be put into 'slope-intercept form'</u>

     ⇒ gotten by isolating the 'y' variable to the left

          y = \frac{3}{4}x-\frac{9}{2}  

<u>Answer:</u>y = \frac{3}{4}x-\frac{9}{2} or y+3=\frac{3}{4} (x-2)

   *<em>Either equations work, put the one that you are the most familiar with</em>

Hope that helps!

#LearnwithBrainly

5 0
2 years ago
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