# Thread: logarithm problem

1. ## logarithm problem

Log3x=y=log9(2x-1)

where the 3 and 9 is subsript

i reach to this point
log3x=log9(2x-1)
log3x=log3(2x-1)
log39
where all the 3 is subscript.
PS i dont understand how log3(2x-1) is derived from log9(2x-1)
log39

where 3 is subscript

2. ## re: logarithm problem

Hello, mathkid12!

$\displaystyle \log_3x \:=\:y\:=\:\log_9(2x-1)$

$\displaystyle \text{i reached this point: }\:\log_3x\:=\:\log_9(2x-1)$

. . $\displaystyle \log_3x \:=\:\frac{\log_3(2x-1)}{\log_39}$

They used the Base Change Formula: .$\displaystyle \log_b(a) \;=\;\frac{\log_n(a)}{\log_n(b)}$

3. ## re: logarithm problem

What would be the end answer please and your help is greatly appreciated

4. ## re: logarithm problem

To continue on what you had:
$\displaystyle \log_3(x)=\frac{\log_3(2x-1)}{\log_3(9)}$
$\displaystyle \Leftrightarrow \log_3(x)=\frac{\log_3(2x-1)}{2}$
$\displaystyle \Leftrightarrow 2\log_3(x)=\log_3(2x-1)$
$\displaystyle \Leftrightarrow \log_3(x^2)=\log_3(2x-1)$

Now use the fact that $\displaystyle \log_a(u(x))=\log_a(v(x)) \Leftrightarrow u(x)=v(x)$

Thanks

6. ## Re: logarithm problem

Note that $\displaystyle log_9(x)= y$ if and only if [tex]x= 9^y= (3^2)^y= 3^{2y}[/itex] which then says that $\displaystyle log_3(x)= 2y$

### log3 x = y= log9 (2x-1)

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