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3 years ago

Find solution to logarithm.

Mathematics

2 answers:

sveta [45]3 years ago

8 0

$\log5x+\log(x-1)=2\\
D:5x>0 \wedge x-1>0\\
D:x>0 \wedge x>1\\
D:x>1\\
\log5x(x-1)=2\\
10^2=5x^2-5x\\
5x^2-5x-100=0\\
x^2-x-20=0\\
x^2-5x+4x-20=0\\
x(x-5)+4(x-5)=0\\
(x+4)(x-5)=0\\
x=-4 \vee x=5\\
-4\not \in D\Rightarrow x=5
$

MAXImum [283]3 years ago

7 0

$log_pa=b\ \ \ \Leftrightarrow\ \ \ p^b=a\\\\loga+logb=log(a\cdot b)\ \ \ \Rightarrow\ \ \ D:\ a>0\ \ \ and\ \ \ b>0\\ ----------------------------- \\log(5x)+log(x-1)=2\ \ \ \Rightarrow\ \ \ D:\ 5x>0\ \ \ and\ \ \ x-1>0\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x>0\ \ \ and\ \ \ \ \ \ \ \ x>1\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=(1;+\infty)\\\\log[5x\cdot(x-1)]=2\ \ \ \Leftrightarrow\ \ \ 5x(x-1)=10^2$

$5x^2-5x=100\ /:5\\\\x^2-x-20=0\\\\x^2-5x+4x-20=0\\\\x(x-5)+4(x-5)=0\\\\(x-5)(x+4)=0\ \ \ \Leftrightarrow\ \ \ x-5=0\ \ \ or\ \ \ x+4=0\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=5\ \ \ or\ \ \ \ \ \ \ \ x=-4\\\\5\in D;\ \ \ -4\notin D\\\\Ans.\ x=5$

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Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

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2 years ago

0.909 > 0.91<br>true or false <br>I need an answer and fast

NeTakaya

Answer:

false

Step-by-step explanation:

0.909 is less than not greater than

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3 years ago

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No link or bot answer the question

sasho [114]

Answer:

c. Rational

because → whole numbers are positive and Starts from 0 so not whole number

Integers can be negative and positive

Examples of numbers that are not integers are: -1.43, 1 3/4, 3.14, . 09, and 5,643.1.

Irrational number that cannot be written in p/q

but -2.4 written in p/q = -24/100

<h3>so c. Rational Number is correct</h3>

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2 years ago

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Answer:

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absolute value means that the sign in front of the # doesn't matter, you just have to look at the positive value, for example:

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Viktor [21]

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