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Harlamova29_29 [7]
3 years ago
8

Solve for x. (e^x-e^-x)/((e^x+e^-x)=t

Mathematics
2 answers:
Natalka [10]3 years ago
6 0

\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=t\\\\\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=(e^x-e^{-x}):(e^x+e^{-x})=\left(e^x-\dfrac{1}{e^x}\right):\left(e^x+\dfrac{1}{e^x}\right)\\\\=\left(\dfrac{(e^x)^2}{e^x}-\dfrac{1}{e^x}\right):\left(\dfrac{(e^x)^2}{e^x}+\dfrac{1}{e^x}\right)=\dfrac{e^{2x}-1}{e^x}:\dfrac{e^{2x}+1}{e^x}\\\\=\dfrac{e^{2x}-1}{e^x}\cdot\dfrac{e^x}{e^{2x}+1}=\dfrac{e^{2x}-1}{e^{2x}+1}\\\\\text{substitute}\ e^{2x}=s


\text{therefore}\\\\\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=t\to\dfrac{s-1}{s+1}=t\ \ \ \ |\cdot(s+1)\\\\s-1=t(s+1)\\\\s-1=ts+t\ \ \ \ |+1\\\\s=ts+t+1\ \ \ |-ts\\\\s-ts=t+1\\\\s(1-t)=t+1\ \ \ \ |:(1-t)\neq0\\\\s=\dfrac{t+1}{1-t}\to e^{2x}=\dfrac{t+1}{1-t}\ \ \ \ |\ln\\\\\ln e^{2x}=\ln\left(\dfrac{t+1}{1-t}\right)\\\\2x\ln e=\ln\left(\dfrac{t+1}{1-t}\right)\\\\2x=\ln\left(\dfrac{t+1}{1-t}\right)\ \ \ \ |:2\\\\\boxed{x=\dfrac{1}{2}\ln\left(\dfrac{t+1}{1-t}\right)}\\\\\text{The domain:}\\\\t\in(-1;\ 1)


\text{Used:}\\\\a^{-n}=\dfrac{1}{a^n}\\\\(a^n)^m=a^{nm}\\\\\log_ab^n=n\cdot\log_ab\\\\\log_aa=1

SIZIF [17.4K]3 years ago
4 0

Answer:

its d on edge ;)

Step-by-step explanation:

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Please help me answer this question​
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her first mistake was on line 3.

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Consider the vector add = a + b = [13 12 11 10 10] and sub = a - b = [-11 - 6 - 14 8]. Find the vector a and b.
prisoha [69]

Answer:

a = [1   3   5   7   9] and b = [12   9   6   3   1]

Step-by-step explanation:

The addition of two vectors a and b is defined as

a + b = [13   12   11   10   10]                .... (1)

The subtraction of two vectors a and b is defined as

a - b = [-11   - 6   - 1   4    8]                 .... (2)

After adding (1) and (2), we get

(a + b) + (a - b)= [13   12   11   10   10] + [-11   - 6   - 1   4    8]

On simplification we get

2a = [13-11   12-6   11-1   10+4   10+8]

2a = [2   6   10   14   18]

Divide both sides by 2.

a = [1   3   5   7   9]

Substitute the value of vector a in equation (1).

[1   3   5   7   9] + b = [13   12   11   10   10]

Subtract vector a from both sides.

b = [13   12   11   10   10] - [1   3   5   7   9]

On simplification we get

b = [13-1   12-3   11-5   10-7   10-9]

b = [12   9   6   3   1]

Therefore the vectors a and b are defined as a = [1   3   5   7   9] and b = [12   9   6   3   1].

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