Assuming [tex]x[/tex] is real:
[tex]\ln e^{\ln x} = \ln x\cdot\ln e=\ln x[/tex]
[tex]\ln e^{\ln x^2} = \ln x^2\cdot\ln e = \ln x^2 = 2\ln x[/tex]
So
[tex]\ln e^{\ln x}+\ln e^{\ln x^2}=2\ln 8\iff\ln x+2\ln x=3\ln x=2\ln 8[/tex]
[tex]\implies\ln x=\dfrac23\ln 8=\ln 8^{2/3}[/tex]
[tex]\implies x=e^{\ln 8^{2/3}}=8^{2/3}=(8^2)^{1/3}=64^{1/3}=4[/tex]
