Log e 4-log 4 e -2A- TrueB- False

The correct option is A TrueLet 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 i ...

Byju's Answer

Standard XI

Mathematics

Fundamental Laws of Logarithms

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

The correct option is A True
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 $x = l o g_{e} 4 \neq 1$ ]

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Similar questions

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

Q. In

△

ABC,

s i n (\frac{A + B}{2}) = c o s (\frac{C}{2})

Q. State true or false.
In

△ A B C

s i n (\frac{A + B}{2}) = c o s (\frac{C}{2})

$(\frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = (\frac{1 - t a n A}{1 - c o t A})^{2} = t a n^{2} A$

Q. If

y = s i n n x + c o s n x

(x, n real), then

- \sqrt{2} \leq y \leq \sqrt{2}

loge4+log4e>2

The correct option is A TrueLet 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]

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

The correct option is A True
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 $x = l o g_{e} 4 \neq 1$ ]