# Inequality ((17)/(25))^k <= 10^(-5)

Deja Bradshaw 2022-10-13 Answered
Inequality ${\left(\frac{17}{25}\right)}^{k}\le {10}^{-5}$ - Solve for $k$
How can I solve for $k$ the following inequality :
${\left(\frac{17}{25}\right)}^{k}\le {10}^{-5}$
This is what I got so far. By taking ${\mathrm{log}}_{k}$ from both sides I get:
${\mathrm{log}}_{k}{\left(\frac{17}{25}\right)}^{k}\le {\mathrm{log}}_{k}{10}^{-5}$
How can I continue from here?
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## Answers (2)

Kristin Myers
Answered 2022-10-14 Author has 12 answers
Using the natural logarithm instead:
$k\mathrm{ln}\frac{17}{25}\le -5\mathrm{ln}10$
Note that $\mathrm{ln}\frac{17}{25}<0$ since $\frac{17}{25}<1$, so by dividing both sides you'll get the equivalent inequality
$k\ge -5\frac{\mathrm{ln}10}{\mathrm{ln}\frac{17}{25}}=5\frac{\mathrm{ln}10}{\mathrm{ln}\frac{25}{17}}=5{\mathrm{log}}_{\frac{25}{17}}10.$
(where the inequality has been flipped, since as noted above we divide both sides by a negative number).
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tikaj1x
Answered 2022-10-15 Author has 4 answers
${\left(\frac{17}{25}\right)}^{k}\le {10}^{-5}\phantom{\rule{0ex}{0ex}}k\mathrm{ln}\left(\frac{17}{25}\right)\le -5\mathrm{ln}\left(10\right)\phantom{\rule{0ex}{0ex}}k\mathrm{ln}\left(\frac{25}{17}\right)\ge 5\mathrm{ln}\left(10\right)\phantom{\rule{0ex}{0ex}}k\ge \frac{5\mathrm{ln}\left(10\right)}{\mathrm{ln}\left(\frac{25}{17}\right)}\phantom{\rule{0ex}{0ex}}k\ge 29.85$
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