How do you solve $4^{5 n} > 30$ $4^(5n)>30$ ?

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How do you solve $4^{5 n} > 30$ $4^(5n)>30$ ?

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Answer 1 · 2018-05-19T15:17:16

$n > 0.49$ $n>0.49$

Explanation:

By rules of logarithms:

${ln a}^{b} = b ln a$ $lna^b=blna$

Therefore, take the logarithm from both sides of the function and follow the rule. Then proceed to rearrange and simplify to find the solution for $n$ $n$ .

${ln 4}^{5} n > ln 30$ $ln4^5n>ln30$

$5 n ln 4 > ln 30$ $5nln4>ln30$

$5 n > \frac{ln 30}{ln 4}$ $5n>ln30/ln4$

$n > \frac{\frac{ln 30}{ln 4}}{5}$ $n>(ln30/ln4)/5$

$n > 0.49$ $n>0.49$

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How do you solve $4^{5 n} > 30$ $4^(5n)>30$ ?

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