What is the value of arccosh(13/5) ?

The value of arccosh(13/5) is ln(5)

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arccosh(13/5)

arccosh (\frac{13}{5})

ln (5)

Decimal

1.60943 \dots

Solution steps

arccosh (\frac{13}{5})

Use the Hyperbolic identity:

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

= ln \frac{13}{5} + (\frac{13}{5})^{2} - 1

ln \frac{13}{5} + (\frac{13}{5})^{2} - 1 = ln (5)

ln \frac{13}{5} + (\frac{13}{5})^{2} - 1

(\frac{13}{5})^{2} - 1 = \frac{12}{5}

(\frac{13}{5})^{2} - 1

(\frac{13}{5})^{2} = \frac{169}{25}

(\frac{13}{5})^{2}

Apply exponent rule:

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

= \frac{1 3 ^{2}}{5 ^{2}}

1 3^{2} = 169

= \frac{169}{5 ^{2}}

5^{2} = 25

= \frac{169}{25}

= \frac{169}{25} - 1

Join

\frac{169}{25} - 1 : \frac{144}{25}

\frac{169}{25} - 1

Convert element to fraction:

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

= \frac{169}{25} - \frac{1 \cdot 25}{25}

Since the denominators are equal, combine the fractions:

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

= \frac{169 - 1 \cdot 25}{25}

169 - 1 \cdot 25 = 144

169 - 1 \cdot 25

Multiply the numbers:

1 \cdot 25 = 25

= 169 - 25

Subtract the numbers:

169 - 25 = 144

= 144

= \frac{144}{25}

= \frac{144}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{144}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{144}{5}

144 = 12

144

Factor the number:

144 = 1 2^{2}

= 1 2^{2}

Apply radical rule:

1 2^{2} = 12

= 12

= \frac{12}{5}

= ln (\frac{13}{5} + \frac{12}{5})

Join

\frac{13}{5} + \frac{12}{5} : 5

\frac{13}{5} + \frac{12}{5}

Since the denominators are equal, combine the fractions:

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

= \frac{13 + 12}{5}

Add the numbers:

13 + 12 = 25

= \frac{25}{5}

Divide the numbers:

\frac{25}{5} = 5

= 5

= ln (5)

= ln (5)