Algebraic equation solving for X

Question

Solve for x

$1^x + 2^x + 3^x = 6^2 $

I have misconceptions here as I keep linking it to the laws of indices and therefore I do not know how to add $1^x + 2^x ... $ can I get a hint ? Thanks !

Answer 1

$1^x + 2^x + 3^x = 6^2\iff 2^x + 3^x = 35$.

Now, you can compute the left side for a few $x$ like this

$2+3=5$

$2^2+3^2=13$

$2^3+3^3=35$

As we know for greater amounts of $x$ the left side will be greater than $35$, though the only answer is $x=3$.

Answer 2

[I merge here my comment with that of user236182]

First observe that the function $f:\mathbb R\rightarrow\mathbb R$ such that $f(x):=1^x+2^x+3^x$ is always increasing (its derivative being $2^x\log(2)+3^x\log(3)>0$ for every $x\in\mathbb R$), so there exists at most a solution; but $f$ is continuous with $f(0)=3<36$ and $f(10)>36$, so a solution exists (and just one).

Now observe that $6^2=(1+2+3)^2$ and recall that $(1+\ldots + n)^2 = 1^3+\ldots +n^3$.

Hence $6^2=1^3+2^3+3^3$ and so the only solution is $x=3$.