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Solve the equation $\varphi(7^x)=294$

$\varphi$ represents Euler's totient function.

I don't understand the solution: $$\varphi(7^x)=7^x\cdot(1-\frac{1}{7})=7^{x-1}\cdot 6=294$$ $$7^{x-1}=49=7^2$$ $$x-1=2\Rightarrow x=3$$

Why is $\varphi(7^x)=7^x\cdot(1-\frac{1}{7})$?

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    See the definition of the Euler's totient function. Two links: [1](https://en.wikipedia.org/wiki/Euler's_totient_function), [2](http://mathworld.wolfram.com/TotientFunction.html). $$\varphi(p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n})$$ $$=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots \left(1-\frac{1}{p_n}\right)$$ – 2017-01-16

2 Answers 2

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This is just a compact way of expressing that the totient function for $7$, $\phi(7)=6$ (like any prime $p$, $\phi(p)=p-1$), and that for higher powers of $7$, $\phi(7^{k+1}) = 7\cdot\phi(7^k)$.

Thus $\phi(7^n)= 6\cdot 7^{n-1} = \frac 67\cdot 7^n$ which matches your expression.

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Why is $\varphi(7^x)=7^x\cdot(1-\frac{1}{7})$?

Because exactly $\frac17$ of the integers $\in[1,7^x]$ have a common factor (larger than $1$) with $7^x$.

The other $\frac67$ integers $\in[1,7^x]$ are coprime to $7^x$.