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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.

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    $\begingroup$ I think, as long as you have the power rule, the last equation stands on its own. The first step is just the power rule, and the second comes from the definition of logarithm. $\endgroup$ – Henry Swanson Nov 13 '13 at 21:31
  • $\begingroup$ @HenrySwanson, how does it go from this: ${b^{x{{\log }_b}(a)}} = {b^{{{\log }_b}({a^x})}} = {a^x}$ , to this: ${\log _a}(x)\cdot \log{_b}(a) = {\log _b}({a^{{{\log }_a}(x)}}) = {\log _b}(x)$ ? $\endgroup$ – seeker Nov 13 '13 at 21:35
  • $\begingroup$ No idea. The first part seems to be unnecessary. $\endgroup$ – Henry Swanson Nov 13 '13 at 22:05
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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 $$ 3 = \log_4 64 = \frac{\log_2 64}{\log_2 4} = \frac{6}{2} = 3 $$

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  • $\begingroup$ Wow, I was expected to string that together from the terse explanation provided in the book? Crazy! Thanks a lot @WillJagy $\endgroup$ – seeker Nov 13 '13 at 22:04
  • $\begingroup$ @Assad, I don't know what is expected of you. I encourage you to write down my first two lines, which are both directions of "exponentiation and logarithm are inverse functions," with my fourth line, $\log_b (a^x) = x \log_b a,$ and try to repeat the derivation of the final result. I used extra letters, that helps a little. $\endgroup$ – Will Jagy Nov 13 '13 at 22:13
  • $\begingroup$ Thank you for your help, I'd have been stuck on this the whole day otherwise! $\endgroup$ – seeker Nov 13 '13 at 22:17
  • $\begingroup$ A quick question, what do you mean by 'purely algebraic substitution'? $\endgroup$ – seeker Nov 25 '13 at 19:57
  • $\begingroup$ Assad, I believe I was emphasizing that no interpretation was used at that step, I just put in a new value for $t.$ $\endgroup$ – Will Jagy Nov 25 '13 at 20:36
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$aLog\left( x \right) = Log\left( {{x^a}} \right)$ 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.

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