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katovenus [111]
2 years ago
9

Can someone please explain how to do the complex algebraic equation? I couldn't find any answers online

Mathematics
1 answer:
frozen [14]2 years ago
3 0
A complex equation is an equation that involves complex numbers when solving it. For example, <4+3i> is a complex number. The 4 or a part is the real part of the number and the 3i or b part is the imaginary part of the number.
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Please help! Its for my big test tomorrow!
Travka [436]

Answer:

# The solution x = -5

# The solution is x = 1

# The solution is x = 6.4

# The solution is x = 4

# The solution is 1.7427

# The solution is 0.190757

Step-by-step explanation:

* Lets revise some rules of the exponents and the logarithmic equation

# Exponent rules:

1- b^m  ×  b^n  =  b^(m + n) ⇒ in multiplication if they have same base

  we add  the power

2- b^m  ÷  b^n =  b^(m – n) ⇒  in division if they have same base we

   subtract  the power

3- (b^m)^n = b^(mn) ⇒ if we have power over power we multiply

   them

4- a^m × b^m = (ab)^m ⇒ if we multiply different bases with same  

   power then we multiply them ad put over the answer the power

5- b^(-m) = 1/(b^m)  (for all nonzero real numbers b) ⇒ If we have

   negative power we reciprocal the base to get positive power

6- If  a^m  =  a^n  ,  then  m  =  n ⇒ equal bases get equal powers

7- If  a^m  =  b^m  ,  then  a  =  b    or    m  =  0

# Logarithmic rules:

1- log_{a}b=n-----a^{n}=b

2- loga_{1}=0---log_{a}a=1---ln(e)=1

3- log_{a}q+log_{a}p=log_{a}qp

4- log_{a}q-log_{a}p=log_{a}\frac{q}{p}

5- log_{a}q^{n}=nlog_{a}q

* Now lets solve the problems

# 3^{x+1}=9^{x+3}

- Change the base 9 to 3²

∴ 9^{x+3}=3^{2(x+3)}=3^{2x+6}

∴ 3^{x+1}=3^{2x+6}

- Same bases have equal powers

∴ x + 1 = 2x + 6 ⇒ subtract x and 6 from both sides

∴ 1 - 6 = 2x - x

∴ -5 = x

* The solution x = -5

# ㏒(9x - 2) = ㏒(4x + 3)

- If ㏒(a) = ㏒(b), then a = b

∴ 9x - 2 = 4x + 3 ⇒ subtract 4x from both sides and add 2 to both sides

∴ 5x = 5 ⇒ divide both sides by 5

∴ x = 1

* The solution is x = 1

# log_{6}(5x+4)=2

- Use the 1st rule in the logarithmic equation

∴ 6² = 5x + 4

∴ 36 = 5x + 4 ⇒ subtract 4 from both sides

∴ 32 = 5x ⇒ divide both sides by 5

∴ 6.4 = x

* The solution is x = 6.4

# log_{2}x+log_{2}(x-3)=2

- Use the rule 3 in the logarithmic equation

∴ log_{2}x(x-3)=2

- Use the 1st rule in the logarithmic equation

∴ 2² = x(x - 3) ⇒ simplify

∴ 4 = x² - 3x ⇒ subtract 4 from both sides

∴ x² - 3x - 4 = 0 ⇒ factorize it into two brackets

∴ (x - 4)(x + 1) = 0 ⇒ equate each bract by 0

∴ x - 4 = 0 ⇒ add 4 to both sides

∴ x = 4

OR

∵ x + 1 = 0 ⇒ subtract 1 from both sides

∴ x = -1

- We will reject this answer because when we substitute the value

 of x in the given equation we will find log_{2}(-1) and this

 value is undefined, there is no logarithm for negative number

* The solution is x = 4

# log_{4}11.2=x

- You can use the calculator directly to find x

∴ x = 1.7427

* The solution is 1.7427

# 2e^{8x}=9.2 ⇒ divide the both sides by 2

∴ e^{8x}=4.6

- Insert ln for both sides

∴ lne^{8x}=ln(4.6)

- Use the rule ln(e^{n})=nln(e) ⇒ ln(e) = 1

∴ 8x = ln(4.6) ⇒ divide both sides by 8

∴ x = ln(4.6)/8 = 0.190757

* The solution is 0.190757

7 0
3 years ago
Read 2 more answers
(7/3,4/3),(-1/3,2/3)
nevsk [136]
How is this a question?
5 0
3 years ago
rewrite 1/4 books over 1/3 hour as a unit rate. A. 1 1/3 bookss/hour B. 1/12 books/hour C. 3/4 books/hour D. 12 books/hour
Nonamiya [84]

Answer:

The Answer would be option C. 3/4 books per hour

Step-by-step explanation:

You would flip 1/3 to 3/1 which makes it a whole number. then multiply

1/4 and 3.

Therefore your answer is 3/4

I hope this helped you,

8 0
3 years ago
I NEED AN EXPERT ASAP PLEASE
tigry1 [53]
The slope in the scatter plot is representative of the trend line. It shows a positive correlation between the independent variable and the dependent variable. since the slope is positive, the data increases.
5 0
3 years ago
Read 2 more answers
Find the distance from P to l.
ki77a [65]
Equation\ of\ a\ line\ from\ 2\ points (x_1;\ y_1);\ (x_2;\ y_2):\\\\y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\(-4;\ 2)\ and\ (3;-5)\\subtitute:\\\\y-2=\dfrac{-5-2}{3-(-4)}\cdot(x-(-4))\\\\y-2=\dfrac{-7}{7}\cdot(x+4)\\\\y-2=-(x+4)\\\\y-2=-x-4\ \ \ |add\ x\ and\ 4\ to\ both\ sides\\\\x+y+2=0\\-------------------\\

Point-line\ distance\\l:Ax+By+C=0;\ (x_0;\ y_0)\\\\d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\\\\x+y+2=0\to A=1;\ B=1;\ C=2\\(1;\ 2)\to x_0=1;\ y_0=2\\subtitute\\\\d=\dfrac{|1\cdot1+1\cdot2+2|}{\sqrt{1^2+1^2}}=\dfrac{|1+2+2|}{\sqrt2}=\dfrac{|5|}{\sqrt2}=\dfrac{5}{\sqrt2}=\dfrac{5\cdot\sqrt2}{\sqrt2\cdot\sqrt2}\\\\=\boxed{\dfrac{5\sqrt2}{2}}
4 0
3 years ago
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