lim x tends to infinity [ln (x^2 + e^x)]/ln[(x^4 + e^(2x))] ?

Lim_x-infinity  [ln (x²+ e^x)]/[ln ( x⁴+e^x)]

So expanding e^x and e^2x we get;

Lin_x-infinity [ln (x²+1+x+x+x²/2!+.........)]/[ln(x⁴+1+2x+(2x)²/2!........)] 

So taking x^{n common in numerator and denominator we can apply limit and    ln (x^n)/ln(x^n) will be obtained so they can be cancelled and the answer will be 1}