What is the general solution for 1+7sinh(x)=4cosh^2(x) ?

The general solution for 1+7sinh(x)=4cosh^2(x) is x=ln(2),x=ln(1+sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

1+7sinh(x)=4cosh^2(x)

Solution

1 + 7 sinh (x) = 4 cosh^{2} (x)

Solution

x = ln (2), x = ln (1 + 2)

Degrees

x = 39.71440 \dots^{\circ}, x = 50.49898 \dots^{\circ}

Solution steps

1 + 7 sinh (x) = 4 cosh^{2} (x)

Rewrite using trig identities

1 + 7 sinh (x) = 4 cosh^{2} (x)

Use the Hyperbolic identity:

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

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 cosh^{2} (x)

Use the Hyperbolic identity:

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

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 (\frac{e ^{x} + e ^{- x}}{2})^{2}

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 (\frac{e ^{x} + e ^{- x}}{2})^{2}

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 (\frac{e ^{x} + e ^{- x}}{2})^{2} : x = ln (2), x = ln (1 + 2)

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 (\frac{e ^{x} + e ^{- x}}{2})^{2}

Apply exponent rules

1 + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} = 4 (\frac{e ^{x} + e ^{- x}}{2})^{2}

Apply exponent rule:

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

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

1 + 7 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 4 (\frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2})^{2}

1 + 7 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 4 (\frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2})^{2}

Rewrite the equation with

e^{x} = u

1 + 7 \cdot \frac{u - ( u ) ^{- 1}}{2} = 4 (\frac{u + ( u ) ^{- 1}}{2})^{2}

Solve

1 + 7 \cdot \frac{u - u ^{- 1}}{2} = 4 (\frac{u + u ^{- 1}}{2})^{2} : u = 2, u = - \frac{1}{2}, u = 1 + 2, u = 1 - 2

1 + 7 \cdot \frac{u - u ^{- 1}}{2} = 4 (\frac{u + u ^{- 1}}{2})^{2}

Refine

1 + \frac{7 ( u ^{2} - 1 )}{2 u} = \frac{( u ^{2} + 1 ) ^{2}}{u ^{2}}

Multiply by LCM

1 + \frac{7 ( u ^{2} - 1 )}{2 u} = \frac{( u ^{2} + 1 ) ^{2}}{u ^{2}}

Find Least Common Multiplier of

2 u, u^{2} : 2 u^{2}

2 u, u^{2}

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

2 u

u^{2}

= 2 u^{2}

Multiply by LCM=

2 u^{2}

1 \cdot 2 u^{2} + \frac{7 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} = \frac{( u ^{2} + 1 ) ^{2}}{u ^{2}} \cdot 2 u^{2}

Simplify

1 \cdot 2 u^{2} + \frac{7 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} = \frac{( u ^{2} + 1 ) ^{2}}{u ^{2}} \cdot 2 u^{2}

Simplify

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

1 \cdot 2 u^{2}

Multiply the numbers:

1 \cdot 2 = 2

= 2 u^{2}

Simplify

\frac{7 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} : 7 u (u^{2} - 1)

\frac{7 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2}

Multiply fractions:

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

= \frac{7 ( u ^{2} - 1 ) \cdot 2 u ^{2}}{2 u}

Cancel the common factor:

2

= \frac{7 ( u ^{2} - 1 ) u ^{2}}{u}

Cancel the common factor:

u

= 7 u (u^{2} - 1)

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

u^{2}

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

2 u^{2} + 7 u (u^{2} - 1) = 2 (u^{2} + 1)^{2}

2 u^{2} + 7 u (u^{2} - 1) = 2 (u^{2} + 1)^{2}

2 u^{2} + 7 u (u^{2} - 1) = 2 (u^{2} + 1)^{2}

Solve

2 u^{2} + 7 u (u^{2} - 1) = 2 (u^{2} + 1)^{2} : u = 2, u = - \frac{1}{2}, u = 1 + 2, u = 1 - 2

2 u^{2} + 7 u (u^{2} - 1) = 2 (u^{2} + 1)^{2}

Expand

2 u^{2} + 7 u (u^{2} - 1) : 2 u^{2} + 7 u^{3} - 7 u

2 u^{2} + 7 u (u^{2} - 1)

Expand

7 u (u^{2} - 1) : 7 u^{3} - 7 u

7 u (u^{2} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 7 u, b = u^{2}, c = 1

= 7 u u^{2} - 7 u \cdot 1

= 7 u^{2} u - 7 \cdot 1 \cdot u

Simplify

7 u^{2} u - 7 \cdot 1 \cdot u : 7 u^{3} - 7 u

7 u^{2} u - 7 \cdot 1 \cdot u

7 u^{2} u = 7 u^{3}

7 u^{2} u

Apply exponent rule:

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

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

= 7 u^{2 + 1}

Add the numbers:

2 + 1 = 3

= 7 u^{3}

7 \cdot 1 \cdot u = 7 u

7 \cdot 1 \cdot u

Multiply the numbers:

7 \cdot 1 = 7

= 7 u

= 7 u^{3} - 7 u

= 7 u^{3} - 7 u

= 2 u^{2} + 7 u^{3} - 7 u

Expand

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

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

(u^{2} + 1)^{2} = u^{4} + 2 u^{2} + 1

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

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u^{2}, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

(u^{2})^{2} = u^{4}

(u^{2})^{2}

Apply exponent rule:

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

= u^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= u^{4}

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

2 u^{2} \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 u^{2}

= u^{4} + 2 u^{2} + 1

= u^{4} + 2 u^{2} + 1

= 2 (u^{4} + 2 u^{2} + 1)

Distribute parentheses

= 2 u^{4} + 2 \cdot 2 u^{2} + 2 \cdot 1

Simplify

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

2 u^{4} + 2 \cdot 2 u^{2} + 2 \cdot 1

Multiply the numbers:

2 \cdot 2 = 4

= 2 u^{4} + 4 u^{2} + 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

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

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

2 u^{2} + 7 u^{3} - 7 u = 2 u^{4} + 4 u^{2} + 2

Switch sides

2 u^{4} + 4 u^{2} + 2 = 2 u^{2} + 7 u^{3} - 7 u

Move

7 u

to the left side

2 u^{4} + 4 u^{2} + 2 = 2 u^{2} + 7 u^{3} - 7 u

Add

7 u

to both sides

2 u^{4} + 4 u^{2} + 2 + 7 u = 2 u^{2} + 7 u^{3} - 7 u + 7 u

Simplify

2 u^{4} + 4 u^{2} + 2 + 7 u = 2 u^{2} + 7 u^{3}

2 u^{4} + 4 u^{2} + 2 + 7 u = 2 u^{2} + 7 u^{3}

Move

7 u^{3}

to the left side

2 u^{4} + 4 u^{2} + 2 + 7 u = 2 u^{2} + 7 u^{3}

Subtract

7 u^{3}

from both sides

2 u^{4} + 4 u^{2} + 2 + 7 u - 7 u^{3} = 2 u^{2} + 7 u^{3} - 7 u^{3}

Simplify

2 u^{4} + 4 u^{2} + 2 + 7 u - 7 u^{3} = 2 u^{2}

2 u^{4} + 4 u^{2} + 2 + 7 u - 7 u^{3} = 2 u^{2}

Move

2 u^{2}

to the left side

2 u^{4} + 4 u^{2} + 2 + 7 u - 7 u^{3} = 2 u^{2}

Subtract

2 u^{2}

from both sides

2 u^{4} + 4 u^{2} + 2 + 7 u - 7 u^{3} - 2 u^{2} = 2 u^{2} - 2 u^{2}

Simplify

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2 = 0

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2 = 0

Factor

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2 : (u - 2) (2 u + 1) (u^{2} - 2 u - 1)

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2

Use the rational root theorem

a_{0} = 2, a_{n} = 2

The dividers of

a_{0} : 1, 2,

The dividers of

a_{n} : 1, 2

Therefore, check the following rational numbers:

\pm \frac{1 , 2}{1 , 2}

\frac{2}{1}

is a root of the expression, so factor out

u - 2

= (u - 2) \frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

\frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} = 2 u^{3} - 3 u^{2} - 4 u - 1

\frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

Divide

\frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} : \frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} = 2 u^{3} + \frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

Divide the leading coefficients of the numerator

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2

and the divisor

u - 2 : \frac{2 u ^{4}}{u} = 2 u^{3}

Quotient = 2 u^{3}

Multiply

u - 2

2 u^{3} : 2 u^{4} - 4 u^{3}

Subtract

2 u^{4} - 4 u^{3}

from

2 u^{4} - 7 u^{3} + 2 u^{2} + 7 u + 2

to get new remainder

Remainder = - 3 u^{3} + 2 u^{2} + 7 u + 2

Therefore

\frac{2 u ^{4} - 7 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} = 2 u^{3} + \frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

= 2 u^{3} + \frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2}

Divide

\frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} : \frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} = - 3 u^{2} + \frac{- 4 u ^{2} + 7 u + 2}{u - 2}

Divide the leading coefficients of the numerator

- 3 u^{3} + 2 u^{2} + 7 u + 2

and the divisor

u - 2 : \frac{- 3 u ^{3}}{u} = - 3 u^{2}

Quotient = - 3 u^{2}

Multiply

u - 2

- 3 u^{2} : - 3 u^{3} + 6 u^{2}

Subtract

- 3 u^{3} + 6 u^{2}

from

- 3 u^{3} + 2 u^{2} + 7 u + 2

to get new remainder

Remainder = - 4 u^{2} + 7 u + 2

Therefore

\frac{- 3 u ^{3} + 2 u ^{2} + 7 u + 2}{u - 2} = - 3 u^{2} + \frac{- 4 u ^{2} + 7 u + 2}{u - 2}

= 2 u^{3} - 3 u^{2} + \frac{- 4 u ^{2} + 7 u + 2}{u - 2}

Divide

\frac{- 4 u ^{2} + 7 u + 2}{u - 2} : \frac{- 4 u ^{2} + 7 u + 2}{u - 2} = - 4 u + \frac{- u + 2}{u - 2}

Divide the leading coefficients of the numerator

- 4 u^{2} + 7 u + 2

and the divisor

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

Quotient = - 4 u

Multiply

u - 2

- 4 u : - 4 u^{2} + 8 u

Subtract

- 4 u^{2} + 8 u

from

- 4 u^{2} + 7 u + 2

to get new remainder

Remainder = - u + 2

Therefore

\frac{- 4 u ^{2} + 7 u + 2}{u - 2} = - 4 u + \frac{- u + 2}{u - 2}

= 2 u^{3} - 3 u^{2} - 4 u + \frac{- u + 2}{u - 2}

Divide

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

Divide the leading coefficients of the numerator

- u + 2

and the divisor

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

Quotient = - 1

Multiply

u - 2

- 1 : - u + 2

Subtract

- u + 2

from

- u + 2

to get new remainder

Remainder = 0

Therefore

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

= 2 u^{3} - 3 u^{2} - 4 u - 1

= 2 u^{3} - 3 u^{2} - 4 u - 1

Factor

2 u^{3} - 3 u^{2} - 4 u - 1 : (2 u + 1) (u^{2} - 2 u - 1)

2 u^{3} - 3 u^{2} - 4 u - 1

Use the rational root theorem

a_{0} = 1, a_{n} = 2

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2}

- \frac{1}{2}

is a root of the expression, so factor out

2 u + 1

= (2 u + 1) \frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1}

\frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1} = u^{2} - 2 u - 1

\frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1}

Divide

\frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1} : \frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1} = u^{2} + \frac{- 4 u ^{2} - 4 u - 1}{2 u + 1}

Divide the leading coefficients of the numerator

2 u^{3} - 3 u^{2} - 4 u - 1

and the divisor

2 u + 1 : \frac{2 u ^{3}}{2 u} = u^{2}

Quotient = u^{2}

Multiply

2 u + 1

u^{2} : 2 u^{3} + u^{2}

Subtract

2 u^{3} + u^{2}

from

2 u^{3} - 3 u^{2} - 4 u - 1

to get new remainder

Remainder = - 4 u^{2} - 4 u - 1

Therefore

\frac{2 u ^{3} - 3 u ^{2} - 4 u - 1}{2 u + 1} = u^{2} + \frac{- 4 u ^{2} - 4 u - 1}{2 u + 1}

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

Divide

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

Divide the leading coefficients of the numerator

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

and the divisor

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

Quotient = - 2 u

Multiply

2 u + 1

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

Subtract

- 4 u^{2} - 2 u

from

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

to get new remainder

Remainder = - 2 u - 1

Therefore

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

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

Divide

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

Divide the leading coefficients of the numerator

- 2 u - 1

and the divisor

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

Quotient = - 1

Multiply

2 u + 1

- 1 : - 2 u - 1

Subtract

- 2 u - 1

from

- 2 u - 1

to get new remainder

Remainder = 0

Therefore

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

= u^{2} - 2 u - 1

= u^{2} - 2 u - 1

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

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

(u - 2) (2 u + 1) (u^{2} - 2 u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u - 2 = 0 or 2 u + 1 = 0 or u^{2} - 2 u - 1 = 0

Solve

u - 2 = 0 : u = 2

u - 2 = 0

Move

2

to the right side

u - 2 = 0

Add

2

to both sides

u - 2 + 2 = 0 + 2

Simplify

u = 2

u = 2

Solve

2 u + 1 = 0 : u = - \frac{1}{2}

2 u + 1 = 0

Move

1

to the right side

2 u + 1 = 0

Subtract

1

from both sides

2 u + 1 - 1 = 0 - 1

Simplify

2 u = - 1

2 u = - 1

Divide both sides by

2

2 u = - 1

Divide both sides by

2

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

Simplify

u = - \frac{1}{2}

u = - \frac{1}{2}

Solve

u^{2} - 2 u - 1 = 0 : u = 1 + 2, u = 1 - 2

u^{2} - 2 u - 1 = 0

Solve with the quadratic formula

u^{2} - 2 u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = - 1

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 1) = 22

(- 2)^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 2^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

u_{1, 2} = \frac{- ( - 2 ) \pm 2 2}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 2}{2 \cdot 1}, u_{2} = \frac{- ( - 2 ) - 2 2}{2 \cdot 1}

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

\frac{- ( - 2 ) + 2 2}{2 \cdot 1}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2

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

\frac{- ( - 2 ) - 2 2}{2 \cdot 1}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

2 - 22 : 2 (1 - 2)

2 - 22

Rewrite as

= 2 \cdot 1 - 22

Factor out common term

2

= 2 (1 - 2)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 - 2

The solutions to the quadratic equation are:

u = 1 + 2, u = 1 - 2

The solutions are

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

1 + 7 \frac{u - u ^{- 1}}{2}

and compare to zero

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

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

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 2 : x = ln (2)

e^{x} = 2

Apply exponent rules

e^{x} = 2

f (x) = g (x)

, then

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

ln (e^{x}) = ln (2)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2)

x = ln (2)

Solve

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

No Solution for

x \in R

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

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

Solve

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

e^{x} = 1 + 2

Apply exponent rules

e^{x} = 1 + 2

f (x) = g (x)

, then

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

ln (e^{x}) = ln (1 + 2)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1 + 2)

x = ln (1 + 2)

Solve

e^{x} = 1 - 2 :

No Solution for

x \in R

e^{x} = 1 - 2

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

x = ln (2), x = ln (1 + 2)

x = ln (2), x = ln (1 + 2)

Graph

Popular Examples

sin^2(x)-1+2cos(2x)-cos^2(x)=0

Frequently Asked Questions (FAQ)

What is the general solution for 1+7sinh(x)=4cosh^2(x) ?
The general solution for 1+7sinh(x)=4cosh^2(x) is x=ln(2),x=ln(1+sqrt(2))