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
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: .$\displaystyle \log_b(a) \;=\;\frac{\log_n(a)}{\log_n(b)}$
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 $\displaystyle \log_a(u(x))=\log_a(v(x)) \Leftrightarrow u(x)=v(x)$