Exponential Equation

I used Newton's method and a programme to solve this question.
Let $f(x)=6^x-5^x-4^x$ .
$f'(x)=(\ln 6)6^x-(\ln 5)5^x-(\ln 4)4^x$
We are finding $x$ for $f(x)=0$
Here we can use iteration $a_{n+1}=a_n-\dfrac{f(a_n)}{f'(a_n)}$ which is the Newton's method.
Pick some initial $x$ for the iteration to converges.
We can obtain the answer is $2.487939\cdots\approx 2.488$

$\text {It can be clearly Seen that for }x\le0 , 6^x<4^x+5^x \\ \text {and For } x\ge4 , 6^x>4^x+5^x \\ \text{we have to check for } x=1,2,3 \\ \text{At }x=1, 4+5=9>6 \\ \text{At }x=2, 16+25=41>36 \\ \text{At }x=3, 64+125=189<216 \\ \text{The inequality has flipped , There must be a Solution between 2 and 3 } \\ \text{At }x=2.5 \text{ we have ,} \\ \implies 2^5+5^{\frac52} \boxed{}6^{\frac52} \\ \implies 2^5+2.236^5\boxed{}2.45^5 \\ \implies 87.9<88.1 \\ \text{From the Options , The Answer will be : } \boxed{2.488}$