logarithm problem

**mathkid12** · December 2nd 2011, 10:10 AM

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

**Soroban** · December 2nd 2011, 10:51 AM

Hello, mathkid12!

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

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

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

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

**mathkid12** · December 2nd 2011, 11:29 AM

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

**Siron** · December 2nd 2011, 11:42 AM

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

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

**mathkid12** · December 4th 2011, 08:16 AM

Thanks

**HallsofIvy** · December 4th 2011, 08:32 AM

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

Thread: logarithm problem

LinkBack

Thread Tools

Display