Confusion regarding the Logarithmic function change of base formula My textbook seems to be making a big leap when trying to prove the change of base formula for logarithms. If someone could help clear this up it would be very appreciated.
It starts with:

$b^{x \log_b(a)}$ 

and uses the power rule to get:

$b^{x \log_b(a)} = b^{\log_b (a^x)}$

And it equates all this to:

$b^{x \log_b(a)} = b^{\log_b(a^x)} = a^x$

Okay, I get it up to here, but then for me it leaps from that to this:

$$\log_a(x)\cdot \log_b(a) = \log_b(a^{\log_a(x)}) = \log_b(x)$$

And it says that divide through by $\log_b(a)$ to get the result.
What precisely has happened here? Could someone walk me through this step-by-step?
Thank you.
 A: I would emphasize these sorts of things: $$ b^{\log_b t} = t  $$ and
$$  \log_b (b^t) = t.  $$
Your third line says $$ b^{x \log_b a} = a^x.  $$ I would take $\log_b$ of both sides to get the one i actually remember
$$ \color{magenta}{ \log_b \left( a^x \right) = x \log_b a }.  $$
That is, to take log with an exponent, pull the exponent in front and apply the log to what remains. 
To match up (eventually) with your text, let me replace theletter $x$ with a letter $t,$ for
$$ \color{red}{ \log_b \left( a^t \right) = t \log_b a }.  $$
Now, make the purely algebraic substitution 
$$    \color{red}{ t = \log_a x  }.  $$
$$ \color{red}{ \log_b \left( a^{\log_a x} \right) = \log_a x \; \; \log_b a }.  $$
However, $$ \color{red}{  a^{\log_a x} = x}.$$
So
 $$ \color{green}{ \log_b \left( x \right) = \log_a x \; \; \log_b a }.  $$
Or
 $$ \color{magenta}{  \log_a x = \; \; \frac{\log_b x}{\log_b a} }.  $$
Notice how the $b$'s seem to cancel in the fraction on the right hand side. Something to help remember. 
For example, with $a=4,b=2,x=64,$ we see
 $$  3 = \log_4 64 =  \frac{\log_2 64}{\log_2 4} = \frac{6}{2} = 3  $$
A: $aLog\left( x \right) = Log\left( {{x^a}} \right)$ according to the rules of logarithms.
Also, ${a^{Lo{g_a}x}} = x$.  I think that this gets you there.  Pretty much what Harry said, but with equations.
