How do you solve ${log}_{3} \sqrt[4]{5} \leq x$ $log_3 root4(5)<=x$ ?

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Answer 1 · 2017-01-07T14:24:23

$x \geq 0.366$ $x>= 0.366$

${log}_{3} \sqrt[4]{5} \leq x$ $log_3 root(4)5 <= x$ can be written as

$x \geq {log}_{3} \sqrt[4]{5}$ $x >= log_3 root(4)5$

or $x \geq {log}_{3} {(5)}^{\frac{1}{4}}$ $x >= log_3 (5)^(1/4)$

or $x \geq \frac{1}{4} {log}_{3} 5$ $x >= 1/4log_3 5$

or $x \geq \frac{1}{4} \frac{log 5}{log 3}$ $x >= 1/4 log5/log3$

or $x \geq \frac{1}{4} \times \frac{0.6990}{0.4771}$ $x >= 1/4xx0.6990/0.4771$

or $x \geq \frac{1}{4} \times 1.465$ $x >= 1/4xx1.465$

or $x \geq 0.366$ $x>= 0.366$

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