solve cosh(x)=2

1) Let's solve in

ℝ

\cosh (x) = 2

Since cosh is even and strictly increasing on

[0, +∞)

, with inverse arccosh,

\cosh (x) = 2

is equivalent to

| x | = arccosh (2)

2) Let's solve in

ℝ

| x | = arccosh (2)

Use the logarithmic form:

arccosh (t) = \ln (t + \sqrt{t^{2} - 1})

ⅰ

arccosh (2)

\ln (2 + \sqrt{2^{2} - 1})

| x | = \ln (2 + \sqrt{2^{2} - 1})

ⓘ Powers are computed before multiplications, additions and subtractions. A power with a positive integer exponent is a repeated product:

2^{2}

2 \cdot 2

4

| x | = \ln (2 + \sqrt{4 - 1})

ⓘ Parentheses are computed first:

4 - 1

3

| x | = \ln (2 + \sqrt{3})

\ln (2 + \sqrt{3}) \geq 0

| x | = \ln (2 + \sqrt{3})

is equivalent to

x = \ln (2 + \sqrt{3})

x = - \ln (2 + \sqrt{3})

▶Solutions found:

x = {- \ln (2 + \sqrt{3}), \ln (2 + \sqrt{3})}

3) The numeric value of:

x = {- \ln (2 + \sqrt{3}), \ln (2 + \sqrt{3})}

is:

x = {- \ln (2 + \approx 1.7321), \ln (2 + \approx 1.7321)}

x = {- \ln (\approx 3.7321), \ln (\approx 3.7321)}

▷

x = {\approx -1.3170, \approx 1.3170}

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