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?

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- Nov 20th 2008, 06:46 AMrtwiltonLog from earlier question, just need confirmation!
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? - Nov 20th 2008, 07:18 AMSoroban

Correct! . . . Nice work!

- Nov 20th 2008, 07:23 AMrtwilton
Thanks! I do have another one here I am working on and could use a jump start. Not sure how to write it. . .

Write the expression 21log3(cube root of x) + log 3(9x^2) - log 3(9) as a single logarithm.... - Nov 20th 2008, 09:15 AMMoo
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 ? - Nov 20th 2008, 09:31 AMrtwilton
I could really use the steps in between. . .I am not making the leap, sorry! I know that subtraction is division, addition is multiplication, correct??

- Nov 20th 2008, 10:17 AMMoo
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 :)

~~~~~~~~~~~~~~~~~~~~~~~~~~

$\displaystyle 21 \log_3 (\sqrt[3]{x})=21 \log_3 (x^{1/3})$

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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)$ - Nov 20th 2008, 10:21 AMrtwilton
Thank you so much, I can see it now. Yes, I believe I will be able to work through future problems, I hope!!