Please solve the inequality.
- Thread starter arbi
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skeeter
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[imath]7^{-\frac{2}{x}}+ 6 \ge 7^{1-\frac{1}{x}}[/imath], note [imath]x \ne 0[/imath]
[imath]7^{-\frac{2}{x}} -1 \ge 7^{1-\frac{1}{x}} -7[/imath]
[imath](7^{-\frac{1}{x}}-1)(7^{-\frac{1}{x}}+1) \ge 7(7^{-\frac{1}{x}}-1)[/imath]
two cases …
1) [imath]7^{-\frac{1}{x}} - 1 >0 \implies 7^{-\frac{1}{x}} \ge 6[/imath]
2) [imath] 7^{-\frac{1}{x}} -1 < 0 \implies 0 < 7^{-\frac{1}{x}} \le 6 [/imath]
proceed …
[imath]7^{-\frac{2}{x}} -1 \ge 7^{1-\frac{1}{x}} -7[/imath]
[imath](7^{-\frac{1}{x}}-1)(7^{-\frac{1}{x}}+1) \ge 7(7^{-\frac{1}{x}}-1)[/imath]
two cases …
1) [imath]7^{-\frac{1}{x}} - 1 >0 \implies 7^{-\frac{1}{x}} \ge 6[/imath]
2) [imath] 7^{-\frac{1}{x}} -1 < 0 \implies 0 < 7^{-\frac{1}{x}} \le 6 [/imath]
proceed …
Last edited:
Steven G
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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)
Last edited:
Otis
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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).
[imath]\;[/imath]