If 0

Question

(B) Let $y = \log_{a} x + \log_{x} a$. Since $x \geq a > 0$ and $a 
eq 1$ (implied by the base of the logarithm),we can use the property $\log_{x} a

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let $y = \log_{a} x + \log_{x} a$. Since $x \geq a > 0$ and $a 
eq 1$ (implied by the base of the logarithm),we can use the property $\log_{x} a = \frac{1}{\log_{a} x}$. Let $u = \log_{a} x$. Since $x \geq a$,$u = \log_{a} x \geq \log_{a} a = 1$. The expression becomes $f(u) = u + \frac{1}{u}$ for $u \geq 1$. By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for $u > 0$,$u + \frac{1}{u} \geq 2\sqrt{u \cdot \frac{1}{u}} = 2$. The equality holds when $u = \frac{1}{u}$,which implies $u^2 = 1$. Since $u \geq 1$,we have $u = 1$. Substituting $u = 1$ back,$\log_{a} x = 1$ implies $x = a$. Thus,the minimum value is $2$.

If $0 < a \leq x$,the minimum value of $\log_{a} x + \log_{x} a$ is

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