How would I show that 1/log(subscript a) b = log(subscript b) a ?
I'm not sure where to start because of the different bases.
Couldbn't I say that
log (subscript b) a = log a / log b is true for any base/subscript.
Then 1 / [log (subscript a) b] = 1 / [log b / log a]
Where 1 / [log b / log a]
=log a / log b
Now use the fact at the top to change this into
log a / log b = log (subscript b) a
Therefore, 1 / [log (subscript a) b] = log (subscript b) a.
?
$\displaystyle a=b^y$
Now substitute in the value we have for $\displaystyle b$, which is $\displaystyle b=a^x$
$\displaystyle a=b^y=\left(a^x\right)^y$ since $\displaystyle b=a^x$
Using the laws of indices... $\displaystyle \left(c^p\right)^q=c^{pq}$
similar to $\displaystyle \left(2^2\right)^3=\left(2^2\right)\left(2^2\right )\left(2^2\right)=2^{2+2+2}=2^6=2^{(2)(3)}$
Therefore $\displaystyle \left(a^x\right)^y=a^{xy}$
but this is equal to $\displaystyle a$
hence $\displaystyle xy=1$
Although this is fine from the perspective of indices, Plato's method is the better one to get used to.
Hello, TN17!
$\displaystyle \text{Show that: }\:\dfrac{1}{\log_a(b)} \:=\: \log_b(a)$
Let the right side equal $\displaystyle \,P\!:\;\;\log_b(a) \:=\:P$ .[1]
Then we have: .$\displaystyle b^P \:=\:a$
Take logs, base $\displaystyle a\!:\;\;\log_a(b^P) \:=\:\underbrace{\log_a(a)}_{\text{This is 1}}$
. . . . . . $\displaystyle P\cdot\log_a(b) \:=\:1 \quad\Rightarrow\quad \log_a(b) \:=\:\dfrac{1}{P}$
Substitute [1]: .$\displaystyle \log_a(b) \;=\;\dfrac{1}{\log_b(a)} $