Changing Logarithm Base: 2log_4(9) - log_2(3)

[Belle]

On this problem, I don't know how to change the base.

2log[sub:26e2h1cd]4[/sub:26e2h1cd]9 - log[sub:26e2h1cd]2[/sub:26e2h1cd]3

I solve it to:

log[sub:26e2h1cd]4[/sub:26e2h1cd]81 - log[sub:26e2h1cd]2[/sub:26e2h1cd]3

But I don't know how to get a common base. Any help would be much appreciated!!

 

[o_O]

Re: Changing Logarithm Base

This rule should help:

\(\displaystyle log_{a} b = \frac{log_{c} b}{log_{c} a}\)

where c can be any workable base: c = 10, e, .. etc:

The next step would be to put it all into one fraction and using the rule again but in reverse to arrive at your single logarithm.

 

[pka]

Re: Changing Logarithm Base

Here is a basic idea that you need.
\(\displaystyle 2\log _4 (9) = 2\frac{{\log (9)}}{{\log (4)}} = 2\frac{{\log (3)}}{{\log (2)}} = 2\log _2 (3)\)

 

[soroban]

Hello, Belle!

\(\displaystyle \text{Simplify: }\;2\log_4(9) - \log_2(3)\)

If we don't know (or forget) the Base Change Formula,
. . . we can still baby-talk our way through it.
(I use this method when I'm tired/confused/desperate.)


\(\displaystyle \text{Let }\;\log_4(9) \:=\\quad\Rightarrow\quad \text{Then: }\;4^p \:=\:9\)

\(\displaystyle \text{Take logs, base 2: }\;\log_2\left(4^p\right) \:=\:\log_2(9)\quad\Rightarrow\quad p\cdot\log_2(4) \:=\:\log_2(9)\)

\(\displaystyle \text{Since }\log_2(4) = 2\text{, we have: }\;p\cdot2 \:=\:\log_2(9)\quad\Rightarrow\quad p \:=\:\frac{1}{2}\log_2(9)\)
. . \(\displaystyle \text{Hence: }\;\log_4(9) \:=\:\frac{1}{2}\log_2(9)\)


\(\displaystyle \text{The expression becomes: }\;2\cdot\frac{1}{2}\log_2(9) - \log_2(3)\)

. . \(\displaystyle =\;\;\log_2(9) - \log_2(3) \;\;=\;\;\log_2\left(\frac{9}{3}\right) \;\;=\;\;\log_2(3)\)

 

[Belle]

Thank you so much for your help!!

That explanation was perfect Soroban!!!