How do you prove $({log}_{a} x) ({log}_{x} a) = 1$ $(log_(a)x)(log_(x)a)=1$ ?

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Answer 1 · 2016-05-06T17:30:20

Please see below.

Let ${log}_{a} x = u$ $log_ax=u$

hence $x = a^{u}$ $x=a^u$ ............(A)

and let ${log}_{x} a = v$ $log_xa=v$ ,

hence $a = x^{v}$ $a=x^v$ ............(B)

Putting value of $a$ $a$ from (B) in (A), we get

$x = {(x^{v})}^{u} = x^{v \times u} = x^{u \times v}$ $x=(x^v)^u=x^(vxxu)=x^(uxxv)$

or $x^{1} = x^{u \times v}$ $x^1=x^(uxxv)$

or $u v = 1$ $uv=1$

Now putting back values of $u$ $u$ and $v$ $v$

${log}_{a} x \times {log}_{x} a = 1$ $log_ax xx log_xa=1$

How do you prove (log_(a)x)(log_(x)a)=1(logax)(logxa)=1(log_(a)x)(log_(x)a)=1?