Inequality $\left(\frac {17}{25}\right)^k \le 10^{-5}$ - Solve for $k$

Question

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 $\log_k$ from both sides I get:

$$ \log_k{\left(\frac {17}{25}\right)^k} \le \log_k{10^{-5}} $$

How can I continue from here?

Answer 1

Using the natural logarithm instead: $$ k \ln \frac{17}{25} \leq -5\ln 10 $$ Note that $\ln \frac{17}{25} < 0$ since $\frac{17}{25} < 1$, so by dividing both sides you'll get the equivalent inequality $$ k \geq -5\frac{\ln 10}{\ln \frac{17}{25}} = 5 \frac{\ln 10}{\ln \frac{25}{17}} = 5\log_{\frac{25}{17}} 10. $$ (where the inequality has been flipped, since as noted above we divide both sides by a negative number).

Answer 2

$$ { \left( \frac { 17 }{ 25 } \right) }^{ k }\le { 10 }^{ -5 }\\ k\ln { \left( \frac { 17 }{ 25 } \right) } \le -5\ln { (10) } \\ k\ln { \left( \frac { 25 }{ 17 } \right) } \ge 5\ln { (10) } \\ k\ge \frac { 5\ln { (10) } }{ \ln { \left( \frac { 25 }{ 17 } \right) } } \\ k\ge 29.85 $$

Answer 3

Take the natural log of both sides; you have $$-k \log(25/17)=k\log(17/25) \le -5 \log(10).$$ Divide and get $$k \ge {5\log(10)\over\log(25/17) }$$