$M(t)=53e^{-t/223}$

let the original size be at $t=0$

$53e^{0} = 53=M_0$

we want to find $t$ such that

$M(t)=\dfrac {M_0}{3} = M_0 e^{-t/223}$

$\dfrac 1 3 = e^{-t/223}$

$\ln(1/3) = -t/223$

$t = -223 \ln(1/3) \approx 245$

For the second problem

$\dfrac {e^x-5}{5e^{-x}-4}=2$

$e^x-5 = 10e^{-x}-8$

$e^x =10e^{-x}-3$

Now a trick. Multiply everything by $e^x$ and get all the terms on one side of the equation.

$e^{2x} + 3e^x -10=0$

Now let $u=e^x$ and we transform this to a quadratic equation in $u$.

$u^2 + 3u - 10=0$

$(u+5)(u-2)=0$

$u=5 \vee u=-2$

$x=\ln(u)$

$\ln(-5)$ is not defined for real numbers so we ignore -5. Thus

$x=\ln(2)$