Having fun with exponents?

Algebra Level 5

$\Large {\left(\sqrt[7]{{\left(\sqrt[7x]{7^x}\right)}^{\sqrt[7x]{7x^7}}}\right)}^{x} = {\left({\left(7^x\right)}^{\sqrt[x]{x}}\right)}^{\left(7^x\right)}$

Find the sum of all real values of $x$ satisfying the real equation above.

Note :- Here $x \neq \{-1,0,1\}$ .

Why don't you have Fun with exponents ?

The answer is -2.00.

1 solution

Ashish Menon
May 11, 2016

$\begin{aligned} \Large {\left(\sqrt[7]{{\left(\sqrt[7x]{7^x}\right)}^{\sqrt[7x]{7x^7}}}\right)}^{x} & = \Large {\left({\left(7^x\right)}^{\sqrt[x]{x}}\right)}^{\left(7^x\right)}\\ \\ \Large {\left({\left({\left({\left(7^x\right)}^{\tfrac{1}{7x}}\right)}^{{\left(7x^7\right)}^{\frac{1}{7x}}}\right)}^{\frac{1}{7}}\right)}^{x} & = \Large {\left({\left(7^x\right)}^{x^{\frac{1}{x}}}\right)}^{7^x}\\ \\ \Large 7^{x × \tfrac{1}{7x} × 7^{\frac{1}{7x}} × {\left(x^7\right)}^{\frac{1}{7x}} × \tfrac{1}{7} × x} & = \Large 7^{x^{\frac{x +1}{x}} × 7^x}\\ \\ \Large 7^{x^{\frac{x + 1}{x}} × 7^{\frac{1 - 14x}{7x}}} & = \Large 7^{x^{\frac{x +1}{x}} × 7^x}\\ \\ \large \text{Equating the powers}:-\\ \Large x^{\tfrac{x + 1}{x}} × 7^{\tfrac{1 - 14x}{7x}} & = \Large x^{\tfrac{x + 1}{x}} × 7^x\\ \\ \Large 7^{\tfrac{1 - 14x}{7x}} & = \Large 7^x\\ \\ \large \text{Equating the powers}:-\\ \large \dfrac{1 - 14x}{7x} & = \large x\\ \\ \large 1 - 14x & = \large 7x^2\\ \\ \large 7x^2 + 14x - 1 & = 0\\ \\ \large \text{So, sum of roots} & = \large -\dfrac{b}{a}\\ & = \large -\dfrac{14}{7}\\ & = \large \boxed{-2} \end{aligned}$

Corrections:

Rows 4 and 5: $\Large 7^{\frac{1-14x}{7x}}$

And Rows 5 to 6, you can't just divide the equation by $\Large x^{\frac{x+1}{x}}$

You should factorize it, and show that there is no solution, like this:

$\Large x^{\frac{x+1}{x}} \times 7^{\frac{1-14x}{7x}} = x^{\frac{x+1}{x}} \times 7^x\\ \Large x^{\frac{x+1}{x}} \times 7^{\frac{1-14x}{7x}} - x^{\frac{x+1}{x}} \times 7^x = 0\\ \Large x^{\frac{x+1}{x}} \left(7^{\frac{1-14x}{7x}} - 7^x \right) = 0\\ \Large x^{\frac{x+1}{x}} = 0,\; 7^{\frac{1-14x}{7x}} - 7^x = 0$

Now, we need to check each condition:

$\Large x^{\frac{x+1}{x}} = 0$

We know that for any $a \neq 0$

$a^b \neq 0$ for all values of $b$

Therefore, as long as $x \neq 0$

$\Large x^{\frac{x+1}{x}} \neq 0$

Since we know $x$ cannot be $0$ , there is no solution for this factor.

The only solution lies in $\Large 7^{\frac{1-14x}{7x}} - 7^x = 0$

Hung Woei Neoh - 5 years, 1 month ago

Yeah I applied what you said but lazy to write thats why I didnt $\text{write}$ dividing $x^{\tfrac{x + 1}{x}}$ on both sides ;) Thanks for the explanation!

Ashish Menon - 5 years, 1 month ago

Ok, I did this process, but then a thought came to my mind: are roots of negative numbers defined? Because if you find each x, you see one is (-14 + sqrt(224))/14 (which is positive) and the other is (-14 - sqrt(224))/14 (which is negative), so xth root in the second case would be a negative root I'm confused about this now

Geovane Coelho - 4 years, 9 months ago

Nice question. I just missed the x in rhs that was over 7.

Puneet Pinku - 5 years, 1 month ago

That means to gave a great try congo :+1:

Ashish Menon - 5 years, 1 month ago

nice problem + nice solution .. +1

Sabhrant Sachan - 5 years, 1 month ago

@Sabhrant Sachan – Thanks! :)

Ashish Menon - 5 years, 1 month ago

Nicely done I did the same way.+1

Rishabh Tiwari - 5 years, 1 month ago

Great :+1:

Ashish Menon - 5 years, 1 month ago

Having fun with exponents?

Why don't you have Fun with exponents ?

The answer is -2.00.

1 solution

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