Let x=log_e 4 so that x gt; 0. Then log_4 e = 1/x. Then x+ 1/x = ( √(x) 1/√(x) )^2 + 2 gt; 2. [Note that we get the strict inequality since x = log_e 4 ...

log e 4+log 4...

Question

$l o g_{e} 4 + l o g_{4} e > 2$

True

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False

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Solution

Let $x = l o g_{e} 4$ so that x > 0. Then $l o g_{4} e = \frac{1}{x}$ .
Then $x + \frac{1}{x} = {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} + 2 > 2$ .
[Note that we get the strict inequality since $x = l o g_{e} 4 \neq 1$ ]

Suggest Corrections

loge4+log4e>2

Let x=loge 4 so that x > 0. Then log4 e=1x. Then x+1x=(√x−1√x)2+2>2. [Note that we get the strict inequality since x=loge 4≠1]

$l o g_{e} 4 + l o g_{4} e > 2$

Let $x = l o g_{e} 4$ so that x > 0. Then $l o g_{4} e = \frac{1}{x}$ .
Then $x + \frac{1}{x} = {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} + 2 > 2$ .
[Note that we get the strict inequality since $x = l o g_{e} 4 \neq 1$ ]