Question

Resolva as equacoes seguintes:

A) $Sinh^{-1}x=2.cosh^{-1}x$

B) $Tanh^{-1}=ln(2x+1)$

Answer 1

Sabendo que:
$\displaystyle \sinh(x)=\frac{e^x-e^{-x}}{2}\implies \sinh^{-1}x=csch(x)=\frac{2}{e^{x}-e^{-x}}\\\\\cosh(x)=\frac{e^{x}+e^{-x}}{2}\implies \cosh^{-1}x=sech(x)=\frac{2}{e^{x}+e^{-x}}$

a)
$\displaystyle i)~~~~\sinh^{-1}x=2\cosh^{-1}x\implies \frac{2}{e^{x}-e^{-x}}=\frac{4}{e^{x}+e^{-x}}\\\\ii)~~~\frac{2e^{x}+2e^{-x}}{e^x-e^{-x}}=4\\\\iii)~~\boxed{\boxed{\frac{e^{x}+e^{-x}}{e^x-e^{-x}}=2}}$

b)
$\displaystyle \tanh^{-1}x=\ln|2x+1|\\\\\text{mas:}~\tanh x=\frac{\sinh x}{\cosh x}=\frac{\frac{e^x-e^{-x}}{2}}{\frac{e^{x}+e^{-x}}{2}}=\frac{e^x-e^{-x}}{e^x+e^{-x}}\implies\\\\\tanh^{-1}x=\frac{e^x+e^{-x}}{e^x-e^{-x}}\\\\i)~~~~\tanh^{-1}x=\ln(2x+1)\\\\ii)~~~\frac{e^x+e^{-x}}{e^x-e^{-x}}=\ln 2x+1\\\\iii)~e^{x}+e^{-x}=(e^x-e^{-x})\ln2x+1\\\\iv)~~e^x+e^{-x}=\ln\left((2x+1)^{e^x-e^{-x}}\right)\\\\v)~~e^{x}=\boxed{\boxed{\ln\left((2x+1)^{e^x-e^{-x}}\right)-e^{-x}}}$

Se trabalhar com a exponencial e o logaritmo é possível simplificar ainda mais essas equações.
Caso tenha problemas para visualizar a resolução, acessar o Brainly através de um navegador da internet.