What is the general solution for (1+tanh(x))/(1-tanh(x))=2 ?

The general solution for (1+tanh(x))/(1-tanh(x))=2 is x= 1/2 ln(2)

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(1+tanh(x))/(1-tanh(x))=2

Solution

\frac{1 + tanh ( x )}{1 - tanh ( x )} = 2

Solution

x = \frac{1}{2} ln (2)

Degrees

x = 19.85720 \dots^{\circ}

Solution steps

\frac{1 + tanh ( x )}{1 - tanh ( x )} = 2

Rewrite using trig identities

\frac{1 + tanh ( x )}{1 - tanh ( x )} = 2

Use the Hyperbolic identity:

tanh (x) = \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}

\frac{1 + \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}}{1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}} = 2

\frac{1 + \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}}{1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}} = 2

\frac{1 + \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}}{1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}} = 2 : x = \frac{1}{2} ln (2)

\frac{1 + \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}}{1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}} = 2

Multiply both sides by

1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}

\frac{1 + \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}}{1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}} (1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}) = 2 (1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})

Simplify

1 + \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = 2 (1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})

Apply exponent rules

1 + \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} = 2 (1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})

Apply exponent rule:

a^{b c} = (a^{b})^{c}

e^{- x} = (e^{x})^{- 1}

1 + \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{e ^{x} + ( e ^{x} ) ^{- 1}} = 2 (1 - \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{e ^{x} + ( e ^{x} ) ^{- 1}})

1 + \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{e ^{x} + ( e ^{x} ) ^{- 1}} = 2 (1 - \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{e ^{x} + ( e ^{x} ) ^{- 1}})

Rewrite the equation with

e^{x} = u

1 + \frac{u - ( u ) ^{- 1}}{u + ( u ) ^{- 1}} = 2 (1 - \frac{u - ( u ) ^{- 1}}{u + ( u ) ^{- 1}})

Solve

1 + \frac{u - u ^{- 1}}{u + u ^{- 1}} = 2 (1 - \frac{u - u ^{- 1}}{u + u ^{- 1}}) : u = 2, u = - 2

1 + \frac{u - u ^{- 1}}{u + u ^{- 1}} = 2 (1 - \frac{u - u ^{- 1}}{u + u ^{- 1}})

Refine

1 + \frac{u ^{2} - 1}{u ^{2} + 1} = 2 (1 - \frac{u ^{2} - 1}{u ^{2} + 1})

Multiply both sides by

u^{2} + 1

1 + \frac{u ^{2} - 1}{u ^{2} + 1} = 2 (1 - \frac{u ^{2} - 1}{u ^{2} + 1})

Multiply both sides by

u^{2} + 1

1 \cdot (u^{2} + 1) + \frac{u ^{2} - 1}{u ^{2} + 1} (u^{2} + 1) = 2 (1 - \frac{u ^{2} - 1}{u ^{2} + 1}) (u^{2} + 1)

Simplify

1 \cdot (u^{2} + 1) + \frac{u ^{2} - 1}{u ^{2} + 1} (u^{2} + 1) = 2 (1 - \frac{u ^{2} - 1}{u ^{2} + 1}) (u^{2} + 1)

Simplify

1 \cdot (u^{2} + 1) : u^{2} + 1

1 \cdot (u^{2} + 1)

Multiply:

1 \cdot (u^{2} + 1) = (u^{2} + 1)

= (u^{2} + 1)

Remove parentheses:

(a) = a

= u^{2} + 1

Simplify

\frac{u ^{2} - 1}{u ^{2} + 1} (u^{2} + 1) : u^{2} - 1

\frac{u ^{2} - 1}{u ^{2} + 1} (u^{2} + 1)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( u ^{2} - 1 ) ( u ^{2} + 1 )}{u ^{2} + 1}

Cancel the common factor:

u^{2} + 1

= u^{2} - 1

u^{2} + 1 + u^{2} - 1 = 2 (1 - \frac{u ^{2} - 1}{u ^{2} + 1}) (u^{2} + 1)

Simplify

u^{2} + 1 + u^{2} - 1 : 2 u^{2}

u^{2} + 1 + u^{2} - 1

Group like terms

= u^{2} + u^{2} + 1 - 1

Add similar elements:

u^{2} + u^{2} = 2 u^{2}

= 2 u^{2} + 1 - 1

1 - 1 = 0

= 2 u^{2}

2 u^{2} = 2 (1 - \frac{u ^{2} - 1}{u ^{2} + 1}) (u^{2} + 1)

2 u^{2} = 2 (1 - \frac{u ^{2} - 1}{u ^{2} + 1}) (u^{2} + 1)

2 u^{2} = 2 (1 - \frac{u ^{2} - 1}{u ^{2} + 1}) (u^{2} + 1)

Expand

2 (1 - \frac{u ^{2} - 1}{u ^{2} + 1}) (u^{2} + 1) : 4

2 (1 - \frac{u ^{2} - 1}{u ^{2} + 1}) (u^{2} + 1)

Expand

(1 - \frac{u ^{2} - 1}{u ^{2} + 1}) (u^{2} + 1) : 2

(1 - \frac{u ^{2} - 1}{u ^{2} + 1}) (u^{2} + 1)

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = - \frac{u ^{2} - 1}{u ^{2} + 1}, c = u^{2}, d = 1

= 1 \cdot u^{2} + 1 \cdot 1 + (- \frac{u ^{2} - 1}{u ^{2} + 1}) u^{2} + (- \frac{u ^{2} - 1}{u ^{2} + 1}) \cdot 1

Apply minus-plus rules

+ (- a) = - a

= 1 \cdot u^{2} + 1 \cdot 1 - \frac{u ^{2} - 1}{u ^{2} + 1} u^{2} - 1 \cdot \frac{u ^{2} - 1}{u ^{2} + 1}

Simplify

1 \cdot u^{2} + 1 \cdot 1 - \frac{u ^{2} - 1}{u ^{2} + 1} u^{2} - 1 \cdot \frac{u ^{2} - 1}{u ^{2} + 1} : 2

1 \cdot u^{2} + 1 \cdot 1 - \frac{u ^{2} - 1}{u ^{2} + 1} u^{2} - 1 \cdot \frac{u ^{2} - 1}{u ^{2} + 1}

1 \cdot u^{2} = u^{2}

1 \cdot u^{2}

Multiply:

1 \cdot u^{2} = u^{2}

= u^{2}

1 \cdot 1 = 1

1 \cdot 1

Multiply the numbers:

1 \cdot 1 = 1

= 1

\frac{u ^{2} - 1}{u ^{2} + 1} u^{2} = \frac{u ^{4} - u ^{2}}{u ^{2} + 1}

\frac{u ^{2} - 1}{u ^{2} + 1} u^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( u ^{2} - 1 ) u ^{2}}{u ^{2} + 1}

Expand

(u^{2} - 1) u^{2} : u^{4} - u^{2}

(u^{2} - 1) u^{2}

= u^{2} (u^{2} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = u^{2}, b = u^{2}, c = 1

= u^{2} u^{2} - u^{2} \cdot 1

= u^{2} u^{2} - 1 \cdot u^{2}

Simplify

u^{2} u^{2} - 1 \cdot u^{2} : u^{4} - u^{2}

u^{2} u^{2} - 1 \cdot u^{2}

u^{2} u^{2} = u^{4}

u^{2} u^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

Add the numbers:

2 + 2 = 4

= u^{4}

1 \cdot u^{2} = u^{2}

1 \cdot u^{2}

Multiply:

1 \cdot u^{2} = u^{2}

= u^{2}

= u^{4} - u^{2}

= u^{4} - u^{2}

= \frac{u ^{4} - u ^{2}}{u ^{2} + 1}

1 \cdot \frac{u ^{2} - 1}{u ^{2} + 1} = \frac{u ^{2} - 1}{u ^{2} + 1}

1 \cdot \frac{u ^{2} - 1}{u ^{2} + 1}

Multiply:

1 \cdot \frac{u ^{2} - 1}{u ^{2} + 1} = \frac{u ^{2} - 1}{u ^{2} + 1}

= \frac{u ^{2} - 1}{u ^{2} + 1}

= u^{2} + 1 - \frac{u ^{4} - u ^{2}}{u ^{2} + 1} - \frac{u ^{2} - 1}{u ^{2} + 1}

Combine the fractions

- \frac{u ^{4} - u ^{2}}{u ^{2} + 1} - \frac{u ^{2} - 1}{u ^{2} + 1} : - u^{2} + 1

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- ( u ^{4} - u ^{2} ) - ( u ^{2} - 1 )}{u ^{2} + 1}

Factor

- (u^{2} - 1) - (u^{4} - u^{2}) : - (u^{2} - 1) (u^{2} + 1)

- (u^{2} - 1) - (u^{4} - u^{2})

Factor

u^{4} - u^{2} : u^{2} (u^{2} - 1)

u^{4} - u^{2}

Factor out common term

u^{2} : u^{2} (u^{2} - 1)

u^{4} - u^{2}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{4} = u^{2} u^{2}

= u^{2} u^{2} - u^{2}

Factor out common term

u^{2}

= u^{2} (u^{2} - 1)

= u^{2} (u^{2} - 1)

= - u^{2} (u^{2} - 1) - (u^{2} - 1)

Factor out common term

(u^{2} - 1)

= (u^{2} - 1) (- u^{2} - 1)

Factor

- u^{2} - 1 : - (u^{2} + 1)

- u^{2} - 1

Factor out common term

- 1

= - (u^{2} + 1)

= - (u^{2} - 1) (u^{2} + 1)

= - \frac{( u ^{2} - 1 ) ( u ^{2} + 1 )}{u ^{2} + 1}

Cancel the common factor:

u^{2} + 1

= - (u^{2} - 1)

Negate

- (u^{2} - 1) = - u^{2} + 1

= - u^{2} + 1

= u^{2} + 1 - u^{2} + 1

Group like terms

= u^{2} - u^{2} + 1 + 1

Add similar elements:

u^{2} - u^{2} = 0

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

= 2

= 2 \cdot 2

Expand

2 \cdot 2 : 4

2 \cdot 2

Distribute parentheses

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

= 4

2 u^{2} = 4

Solve

2 u^{2} = 4 : u = 2, u = - 2

2 u^{2} = 4

Divide both sides by

2

2 u^{2} = 4

Divide both sides by

2

\frac{2 u ^{2}}{2} = \frac{4}{2}

Simplify

u^{2} = 2

u^{2} = 2

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 2, u = - 2

u = 2, u = - 2

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

1 + \frac{u - u ^{- 1}}{u + u ^{- 1}}

and compare to zero

u = 0

Take the denominator(s) of

2 (1 - \frac{u - u ^{- 1}}{u + u ^{- 1}})

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 2, u = - 2

u = 2, u = - 2

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 2 : x = \frac{1}{2} ln (2)

e^{x} = 2

Apply exponent rules

e^{x} = 2

Apply exponent rule:

a = a^{\frac{1}{2}}

2 = 2^{\frac{1}{2}}

e^{x} = 2^{\frac{1}{2}}

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (2^{\frac{1}{2}})

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2^{\frac{1}{2}})

Apply log rule:

ln (x^{a}) = a \cdot ln (x)

ln (2^{\frac{1}{2}}) = \frac{1}{2} ln (2)

x = \frac{1}{2} ln (2)

x = \frac{1}{2} ln (2)

Solve

e^{x} = - 2 :

No Solution for

x \in R

e^{x} = - 2

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

x = \frac{1}{2} ln (2)

Verify Solutions:

x = \frac{1}{2} ln (2)

True

Check the solutions by plugging them into

\frac{1 + \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}}{1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}} = 2

Remove the ones that don't agree with the equation.

Plug in

x = \frac{1}{2} ln (2) :

True

\frac{1 + \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}}{1 - \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}} = 2

\frac{1 + \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}}{1 - \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}} = 2

\frac{1 + \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}}{1 - \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}}

\frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}} = \frac{1}{3}

\frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}

e^{\frac{1}{2} l n (2)} = 2

e^{\frac{1}{2} l n (2)}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

= e^{l n (2)}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

= 2

e^{- \frac{1}{2} l n (2)} = \frac{1}{2}

e^{- \frac{1}{2} l n (2)}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

= (e^{l n (2)})^{- \frac{1}{2}}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

= 2^{- \frac{1}{2}}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{2}

= \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{2 + \frac{1}{2}}

e^{\frac{1}{2} l n (2)} = 2

e^{\frac{1}{2} l n (2)}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

= e^{l n (2)}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

= 2

e^{- \frac{1}{2} l n (2)} = \frac{1}{2}

e^{- \frac{1}{2} l n (2)}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

= (e^{l n (2)})^{- \frac{1}{2}}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

= 2^{- \frac{1}{2}}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{2}

= \frac{2 - \frac{1}{2}}{2 + \frac{1}{2}}

Join

2 + \frac{1}{2} : \frac{3}{2}

2 + \frac{1}{2}

Convert element to fraction:

2 = \frac{2 2}{2}

= \frac{2 2}{2} + \frac{1}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 2 + 1}{2}

22 + 1 = 3

22 + 1

Apply radical rule:

a a = a

22 = 2

= 2 + 1

Add the numbers:

2 + 1 = 3

= 3

= \frac{3}{2}

= \frac{2 - \frac{1}{2}}{\frac{3}{2}}

Join

2 - \frac{1}{2} : \frac{1}{2}

2 - \frac{1}{2}

Convert element to fraction:

2 = \frac{2 2}{2}

= \frac{2 2}{2} - \frac{1}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 2 - 1}{2}

22 - 1 = 1

22 - 1

Apply radical rule:

a a = a

22 = 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= \frac{1}{2}

= \frac{\frac{1}{2}}{\frac{3}{2}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{1 \cdot 2}{2 \cdot 3}

Refine

= \frac{2}{2 \cdot 3}

Cancel the common factor:

2

= \frac{1}{3}

= \frac{1 + \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}}{1 - \frac{1}{3}}

\frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}} = \frac{1}{3}

\frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}

e^{\frac{1}{2} l n (2)} = 2

e^{\frac{1}{2} l n (2)}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

= e^{l n (2)}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

= 2

e^{- \frac{1}{2} l n (2)} = \frac{1}{2}

e^{- \frac{1}{2} l n (2)}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

= (e^{l n (2)})^{- \frac{1}{2}}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

= 2^{- \frac{1}{2}}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{2}

= \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{2 + \frac{1}{2}}

e^{\frac{1}{2} l n (2)} = 2

e^{\frac{1}{2} l n (2)}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

= e^{l n (2)}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

= 2

e^{- \frac{1}{2} l n (2)} = \frac{1}{2}

e^{- \frac{1}{2} l n (2)}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

= (e^{l n (2)})^{- \frac{1}{2}}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

= 2^{- \frac{1}{2}}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{2}

= \frac{2 - \frac{1}{2}}{2 + \frac{1}{2}}

Join

2 + \frac{1}{2} : \frac{3}{2}

2 + \frac{1}{2}

Convert element to fraction:

2 = \frac{2 2}{2}

= \frac{2 2}{2} + \frac{1}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 2 + 1}{2}

22 + 1 = 3

22 + 1

Apply radical rule:

a a = a

22 = 2

= 2 + 1

Add the numbers:

2 + 1 = 3

= 3

= \frac{3}{2}

= \frac{2 - \frac{1}{2}}{\frac{3}{2}}

Join

2 - \frac{1}{2} : \frac{1}{2}

2 - \frac{1}{2}

Convert element to fraction:

2 = \frac{2 2}{2}

= \frac{2 2}{2} - \frac{1}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 2 - 1}{2}

22 - 1 = 1

22 - 1

Apply radical rule:

a a = a

22 = 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= \frac{1}{2}

= \frac{\frac{1}{2}}{\frac{3}{2}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{1 \cdot 2}{2 \cdot 3}

Refine

= \frac{2}{2 \cdot 3}

Cancel the common factor:

2

= \frac{1}{3}

= \frac{1 + \frac{1}{3}}{1 - \frac{1}{3}}

Simplify

\frac{1 + \frac{1}{3}}{1 - \frac{1}{3}}

Join

1 - \frac{1}{3} : \frac{2}{3}

1 - \frac{1}{3}

Convert element to fraction:

1 = \frac{1 \cdot 3}{3}

= \frac{1 \cdot 3}{3} - \frac{1}{3}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot 3 - 1}{3}

1 \cdot 3 - 1 = 2

1 \cdot 3 - 1

Multiply the numbers:

1 \cdot 3 = 3

= 3 - 1

Subtract the numbers:

3 - 1 = 2

= 2

= \frac{2}{3}

= \frac{1 + \frac{1}{3}}{\frac{2}{3}}

Join

1 + \frac{1}{3} : \frac{4}{3}

1 + \frac{1}{3}

Convert element to fraction:

1 = \frac{1 \cdot 3}{3}

= \frac{1 \cdot 3}{3} + \frac{1}{3}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot 3 + 1}{3}

1 \cdot 3 + 1 = 4

1 \cdot 3 + 1

Multiply the numbers:

1 \cdot 3 = 3

= 3 + 1

Add the numbers:

3 + 1 = 4

= 4

= \frac{4}{3}

= \frac{\frac{4}{3}}{\frac{2}{3}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{4 \cdot 3}{3 \cdot 2}

Cancel the common factor:

3

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

= 2

2 = 2

True

The solution is

x = \frac{1}{2} ln (2)

x = \frac{1}{2} ln (2)

Popular Examples

2sin^2(θ)+5sin(θ)-3=0

2 sin^{2} (θ) + 5 sin (θ) - 3 = 0

sin(2x)-sin(x)=0,-pi<= x<= pi

sin (2 x) - sin (x) = 0, - π \leq x \leq π

2(tan(x)+3)=5+tan(x)

2 (tan (x) + 3) = 5 + tan (x)

3cos(4θ)=-2

3 cos (4 θ) = - 2

solvefor x,sin(x)=sin(y)solve for

x, sin (x) = sin (y)

Frequently Asked Questions (FAQ)

What is the general solution for (1+tanh(x))/(1-tanh(x))=2 ?
The general solution for (1+tanh(x))/(1-tanh(x))=2 is x= 1/2 ln(2)