What is the value of arccosh(5/3) ?

The value of arccosh(5/3) is ln(3)

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arccosh(5/3)

arccosh (\frac{5}{3})

ln (3)

Decimal Notation

1.09861 \dots

Solution steps

arccosh (\frac{5}{3})

Use the Hyperbolic identity:

arccosh (x) = ln (x + x^{2} - 1)

= ln \frac{5}{3} + (\frac{5}{3})^{2} - 1

ln \frac{5}{3} + (\frac{5}{3})^{2} - 1 = ln (3)

ln \frac{5}{3} + (\frac{5}{3})^{2} - 1

(\frac{5}{3})^{2} - 1 = \frac{4}{3}

(\frac{5}{3})^{2} - 1

(\frac{5}{3})^{2} = \frac{25}{9}

(\frac{5}{3})^{2}

Apply exponent rule:

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

= \frac{5 ^{2}}{3 ^{2}}

5^{2} = 25

= \frac{25}{3 ^{2}}

3^{2} = 9

= \frac{25}{9}

= \frac{25}{9} - 1

Join

\frac{25}{9} - 1 : \frac{16}{9}

\frac{25}{9} - 1

Convert element to fraction:

1 = \frac{1 \cdot 9}{9}

= \frac{25}{9} - \frac{1 \cdot 9}{9}

Since the denominators are equal, combine the fractions:

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

= \frac{25 - 1 \cdot 9}{9}

25 - 1 \cdot 9 = 16

25 - 1 \cdot 9

Multiply the numbers:

1 \cdot 9 = 9

= 25 - 9

Subtract the numbers:

25 - 9 = 16

= 16

= \frac{16}{9}

= \frac{16}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{16}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{16}{3}

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= \frac{4}{3}

= ln (\frac{5}{3} + \frac{4}{3})

Join

\frac{5}{3} + \frac{4}{3} : 3

\frac{5}{3} + \frac{4}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{5 + 4}{3}

Add the numbers:

5 + 4 = 9

= \frac{9}{3}

Divide the numbers:

\frac{9}{3} = 3

= 3

= ln (3)

= ln (3)

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