Why is $y{(\log_a(x))} = \log_a{(x^y)}$?

Question

Why is $y{(\log_a(x))} = \log_a{(x^y)}$?

I feel like I'm missing something here. Sorry if I put the title wrong..

Answer 1

It is one of the logarithmic identities: $$\log(a^b) = b\log(a)$$

Answer 2

So, check it out, when $y$ is a positive integer, $x$ is real, let's give a definition of $$ x^y $$ It is of course, defined precisely the way that you think, $$ x^y = \underbrace{x\cdot x\cdot x\cdot \ldots\cdot x}_{y \;\text{times}} $$ So, in particular, for $z,y$ postive integers: $$ (x^{z})^y = \underbrace{(\underbrace{x\cdot x\cdot x\cdot \ldots\cdot x}_{z \;\text{times}})\cdot( \underbrace{x\cdot x\cdot x\cdot \ldots\cdot x}_{z \;\text{times}})\cdot \ldots \cdot (\underbrace{x\cdot x\cdot x\cdot \ldots\cdot x}_{z \;\text{times}})}_{y\; \text{times}} = \underbrace{x\cdot x\cdot x\cdot \ldots\cdot x}_{z\cdot y \; \text{times}} $$ Now, remember that $\log_b(x)$ is the inverse function of $b^x$. So, in particular, $$ b^{ac}=x \longleftrightarrow \log_b(x)=ac $$ Applying the above thoughts: $$ \log_b(x)=\log_b(b^{ac})=\log_b\left(\underbrace{(\underbrace{b\cdot b\cdot b\cdot \ldots\cdot b}_{a \;\text{times}})\cdot( \underbrace{b\cdot b\cdot b\cdot \ldots\cdot b}_{a \;\text{times}})\cdot \ldots \cdot (\underbrace{b\cdot b\cdot b\cdot \ldots\cdot b}_{a \;\text{times}})}_{c\; \text{times}}\right) $$ Now by properties of $\log$, (which we'll get to in a moment), $$ \log_b\left(\underbrace{(\underbrace{b\cdot b\cdot b\cdot \ldots\cdot b}_{a \;\text{times}})\cdot( \underbrace{b\cdot b\cdot b\cdot \ldots\cdot b}_{a \;\text{times}})\cdot \ldots \cdot (\underbrace{b\cdot b\cdot b\cdot \ldots\cdot b}_{a \;\text{times}})}_{c\; \text{times}}\right)=\sum_{1}^{c}\log_b(\underbrace{b\cdot b\cdot b\cdot \ldots\cdot b}_{a\; \text{times}})$$ $$ =ac $$ To see why $\log_b(b^x\cdot b^y) = \log_b(b^x)+\log_b(b^y)$. It is sufficient to convince yourself of the property $b^x\cdot b^y = b^{x+y}$, in a similar fashion as above. The above property allows us to break up products in the arguments of $\log$ into the sums of the $\log$'s of the components of the products