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cestrela7 [59]
3 years ago
13

The value of x in the equation log(x-1)9=2 is

Mathematics
2 answers:
mel-nik [20]3 years ago
4 0
\log_{(x-1)} 9=2

the domain:
x-1 >0 \ \land \ x-1 \not=1 \\
x>1 \ \land \ x \not= 2 \\
x \in (1; 2) \cup (2;+\infty)

the equation:
\log_{(x-1)}9=2 \\
(x-1)^2=9 \\
\sqrt{(x-1)^2}=\sqrt{9} \\
|x-1|=3 \\
x-1=3 \ \lor \ x-1=-3 \\
x=4 \ \lor \ x=-2

4 is in the domain
-2 is not in the domain

The answer:
x=4

***
Also, the answer in 13 is {8}. -3 is not in the domain. Replace x with -3 and you'll see:
x=\sqrt{5x+24} \\
-3=\sqrt{5 \times (-3)+24} \\
-3=\sqrt{-15+24} \\
-3=\sqrt{9} \\
-3=3
It's not true so -3 isn't a solution to this equation.
maxonik [38]3 years ago
3 0
D:x-1>0 \wedge x\n-1\not=1\\
D:x>1 \wedge x\not=2\\
D:x\in(1,2)\cup(2,\infty)\\\\
\log_{x-1}9=2\\
(x-1)^2=9\\
x-1=3 \vee x-1=-3\\
x=4 \vee x=-2\\-2\not \in D\\
\boxed{ x=4}
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3 years ago
Find each of the following for ​
KATRIN_1 [288]
<h2>Answer:</h2>

(a)

          f(x+ h)=8x+8h+3  

(b)

            f(x+ h)-f(x)=8h          

(c)

             \dfrac{f(x+ h)-f(x)}{h}=8

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as :

              f(x)=8x+3

(a)

           f(x+ h)

We will substitute (x+h) in place of x in the function f(x) as follows:

f(x+h)=8(x+h)+3\\\\i.e.\\\\f(x+h)=8x+8h+3

(b)

       f(x+ h)-f(x)              

Now on subtracting the f(x+h) obtained in part (a) with the function f(x) we have:

f(x+h)-f(x)=8x+8h+3-(8x+3)\\\\i.e.\\\\f(x+h)-f(x)=8x+8h+3-8x-3\\\\i.e.\\\\f(x+h)-f(x)=8h

(c)

           \dfrac{f(x+ h)-f(x)}{h}            

In this part we will divide the numerator expression which is obtained in part (b) by h to get:

           \dfrac{f(x+ h)-f(x)}{h}=\dfrac{8h}{h}\\\\i.e.\\\\\dfrac{f(x+h)-f(x)}{h}=8    

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they have different answers

Step-by-step explanation:

X - 13 = 4 and 13 - X = 4

for x- 13 = 4

x=4+13=17

x=17

for 13 - x = 4

13-4-x=0

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9=x

x=9

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