Question

If 0 <a< x, then minimum value of
$| log_a {x + log_x{a} } |$
is ?​

Answer 1

Solution:-

$\implies \rm \: | log_{a}x + log_{x}a |$

Condition

$\rm \: \implies0 < a < x$

Now use the property of AM and GM

$\rm \implies \dfrac{a + b}{2} \geqslant \sqrt{a \times b}$

We can write as

$\rm \: \implies \dfrac{ log_{a}x + log_{x}a}{2} \geqslant \sqrt{ log_{a}x \times log_{x}a }$

Use the logarithmic properties

$\implies \boxed{ \rm log_{m}n = \dfrac{ log \: n }{ log \: m } }$

Now

$\rm\implies \: log_{a}x + log_{x}a \geqslant 2( \sqrt{ log_{a}x \times log_{x}a}$

using this property

$\rm\implies \: log_{a}x + log_{x}a \geqslant 2( \sqrt{ \dfrac{ log \: x }{ log \: a } \times \dfrac{ log \: a }{ log \: x } }$

we get

$\rm\implies \: log_{a}x + log_{x}a \geqslant 2( \sqrt{ \cancel\dfrac{ log \: x }{ log \: a } \times \cancel \dfrac{ log \: a }{ log \: x } }$

$\rm\implies \: log_{a}x + log_{x}a \geqslant 2( 1)$

$\rm\implies \: log_{a}x + log_{x}a \geqslant 2$

So the minimum value of

$\rm\implies \: log_{a}x + log_{x}a \: \: is \: \: 2$

Answer 2

Answer:

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