I have log4(x^2 - 9) - log4(x + 3) = 3
I get 4^3 = (x^2 - 9) / (x + 3 )
and then 64 = (x - 3)
x = 67
Correct?
Use these formulae :
$\displaystyle a \log b =\log b^a$
$\displaystyle \log a + \log b- \log c=\log \left(\frac{ab}{c}\right)$
and this holds true for any base of the logarithm.
note that $\displaystyle \sqrt[3]{x}=x^{1/3}$
So $\displaystyle 21 \log_3 \sqrt[3]{x}+\log_3 (9x^2)-\log_3(9)=\dots=\log_3 \left(\frac{x^7 \cdot 9x^2}{9}\right)=\log_3 (x^9)=9 \log_3(x)$
d'you get all the calculations ?
Yep !
But don't make this mistake :
$\displaystyle \log(a+b) {\color{red} \neq } \log(a) \cdot \log(b)$
Okay, here are the steps, hope you'll know how to do further exercises later
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$\displaystyle 21 \log_3 (\sqrt[3]{x})=21 \log_3 (x^{1/3})$
from the property a log(b)=log(b^a), this is :
$\displaystyle \log_3 \left((x^{1/3})^{21}\right)$
use the rule $\displaystyle (a^b)^c=(a^c)^b=a^{bc}$ and you'll get :
$\displaystyle 21 \log_3 (\sqrt[3]{x})=\log_3 \left(x^{21/3}\right)=\log_3 (x^7)$
this is the main step of the leap
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I've just seen a better way to do it :
$\displaystyle 21 \log_3 (x^{1/3})+\log_3(9x^2)-\log_3(9)=7 \log_3(x)+\log_3(9)+\log_3(x^2)-\log_3(9)$$\displaystyle =7\log_3(x)+2 \log_3(x)=9 \log_3(x)$