This actually triggered me in my mind from here. After some playing around I notice that the relation $a^{\log_b x} = x^{\log_b a}$ is true for any valid value of $a,b$ and $x$. I am very inquisitive to see how this holds ?
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
For any positive $r$ and any $s$, you have $$r^s = b^{\log_b(r^s)} = b^{s\log_b(r)}.$$ So, taking $r=a$ and $s=\log_b(x)$, we have: \begin{align*} a^{\log_b(x)} &= b^{\log_b(x)\log_b(a)}\\ &= b^{\log_b(a)\log_b(x)}\\ & = b^{\log_b(x^{\log_b(a)})}\\ &= x^{\log_b(a)}. \end{align*} |
|||
|
|
|
HINT $\rm\ \ \ log(A^{\log X})\ =\ log\ X\ \ log\ A\ =\ log(X^{\log A})\:,\ $ where $\rm\ log := log_b$ |
|||
|
|
|
Using log properties, we have $a^{\log_b(x)} = b^{\log_b\left(a^{\log_b(x)}\right)} = b^{\log_b(x)\log_b(a)} = \left(b^{\log_b(x)}\right)^{\log_b(a)} = x^{\log_b(a)}$ |
|||
|
|
|
Otherwise : Changing the base of logarithm, we have : $$\log_{b}a= \displaystyle\frac{\log_{x}a}{\log_{x}b}$$ and $$\log_{x}b= \displaystyle\frac{\log_{b}b}{\log_{b}x} = \frac{1}{\log_{b}x} $$ By combine these two ecuations, $\log_{b}a= (\log_{b}x)(\log_{x}a) \Leftrightarrow (\log_{b}a)(\log_{x}x)= (\log_{b}x)(\log_{x}a) \Leftrightarrow \log_{x}x^{\log_{b}a} = \log_{x}a^{\log_{b}x}$ By last, canceling $\log_{x}$ on both sides, we have : $$x^{\log_{b}a} = a^{\log_{b}x}$$ |
|||
|
|
