Please solve the inequality.

[arbi]

Please solve the inequality.
I got the solution [1 / log 7(1/6); + infinity), but not sure if it's true

 

[skeeter]

$7^{-\frac{2}{x}}+ 6 \ge 7^{1-\frac{1}{x}}$, note $x \ne 0$

$7^{-\frac{2}{x}} -1 \ge 7^{1-\frac{1}{x}} -7$

$(7^{-\frac{1}{x}}-1)(7^{-\frac{1}{x}}+1) \ge 7(7^{-\frac{1}{x}}-1)$

two cases …

1) $7^{-\frac{1}{x}} - 1 >0 \implies 7^{-\frac{1}{x}} \ge 6$
2) $7^{-\frac{1}{x}} -1 < 0 \implies 0 < 7^{-\frac{1}{x}} \le 6$

proceed …

 

[Steven G]

Please solve the inequality.

Hi,
No one here is going to solve your problem for you. After all, this is a math help forum where we help students solve their problem.
So please share the work you have done so far.
Here is another hint: Let u = 7^(-1/x)

 

[Otis]

I got the solution [1 / log 7(1/6); + infinity), but not sure if it's true

Hi arbi. You can check your solution, by using a scientific calculator.

Evaluate 1/log_7(1/6), and then substitute the decimal form for x, in the given inequality. Check x-values slightly above and below that value, to confirm.

We can use properties of exponents and logarithms, to simplify expressions like 1/log_7(1/6).

Also, be mindful of domain, when working with variable expressions. Your solution interval contains zero.

PS: Did you see how I typed "log base 7 of 1/6"? Using an underscore to show the base is one way. Another way uses square brackets: log[7](1/6).
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