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I am having a hard time with logarithmic properties, and I was wondering how to expand this equation by using the properties. The equation and the steps I have done so far are $$\ln\left(2b \sqrt\frac{b+1}{b-1}\right)$$ $$\text{Quotient rule:}\phantom2 \ln(2b\sqrt{b+1})-\ln(\sqrt{b-1})$$ $$ \text{Exponent rule:} \phantom2 \frac12\ln(2b(b+1))-\frac12\ln(2b(b-1))$$

Is this right so far? Is this what you would do to expand this equation by using logarithmic properties?

thanks

Gary
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Benp404
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    A mental trick to handle problems like this is to write the argument as a product of $z^k$ terms; each such term becomes a sum-term $k\ln z$. – Parcly Taxel Nov 06 '21 at 11:08
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    What you did with the "exponent rule" is incorrect: $$ \frac{1}{2}\ln (2b(b + 1)) - \frac{1}{2}\ln (2b(b - 1)) = \ln \left( {\frac{{\sqrt {2b(b + 1)} }}{{\sqrt {2b(b - 1)} }}} \right) = \ln \left( {\sqrt {\frac{{b + 1}}{{b - 1}}} } \right) \ne \ln \left( {2b\sqrt {\frac{{b + 1}}{{b - 1}}} } \right). $$ – Gary Nov 06 '21 at 11:16
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    More generally, you can write$$\ln\left(x\sqrt{y/z}\right)=\tfrac12\ln(x^2yz)=\ln x+\tfrac12\ln y-\tfrac12\ln z$$or$$\ln\left(x\sqrt{y/z}\right)=\ln x+\ln\sqrt{y/z}=\ln x+\tfrac12\ln y-\tfrac12\ln z.$$Your quotient rule worked, but your exponent rule was incorrect; it reduced to $\tfrac12\ln y-\tfrac12\ln z$. – J.G. Nov 06 '21 at 11:22

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Your exponent rule is incorrectly applied: $\ln(2b \sqrt{b + 1}) ≠ \frac{1}{2}\ln(2b(b + 1))$. Instead, $\frac{1}{2}\ln(2b(b + 1)) = \ln(\sqrt{(2b(b + 1))})$. Note how the root applies to entire function inside.

$$\to \ln(2b \sqrt{b+1}) - \ln(\sqrt{b-1})$$ $$\to \ln(2b) + \ln(\sqrt{b+1}) - \ln(\sqrt{b-1})$$ $$\to \ln(2b) + \frac{1}{2}(\ln(\frac{b + 1}{b-1}))$$ $$\to \ln(2b) + \frac{1}{2}\ln(1+\frac{2}{b-1})$$

You may be able to expand further. Also logarithmics are funny when it comes to domain and functions. Watch out and check that domain matches to your initial given function, in every step when working with logarithms, as you might introduce extra solutions from rationalising denominators to separating logs.

Dstarred
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  • Thanks for the help. I worked my way through most of the problem. $ln(2b\sqrt\frac{b+1}{b-1})$ $$ln(2)+ln(b)+\frac12ln(b+1)-\frac12ln(b_1)$$ $$ln(2)+ln(b)+\frac{(b+1)-(b-1)}{2}$$ – Benp404 Nov 07 '21 at 01:31
  • The last line is to read: $$ln(2)+ln(b)+\frac{ln(b+1)-ln(b-1}{2}$$, I did still have a question about why the $ln(b+1)-ln(b-1)$ are put over 2. Is it safe to say since both the terms $ln(b+1)$ and $ln(b-1)$ are being multiplied by 1/2 and that is the reason why they two terms are put into the numerator? – Benp404 Nov 07 '21 at 01:44
  • @Benp404 yes both of them are being multiplied by 1/2. Since the terms are also being subtracted, you can simplify it in one fraction, by putting it over 2 (since its half). $$\frac{1}{2} \times \ln{(b + 1)} = \frac{\ln{(b + 1)}}{2}$$ Similarly, applying it to both at once becomes $$\frac{\ln{(b + 1)}}{2} - \frac{\ln{(b + 1)}}{2}$$. You can combine both numerators just like you did after this step if you want just 1 fraction. – Dstarred Nov 07 '21 at 11:49
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    I think the domain is b>1. – Benp404 Nov 08 '21 at 12:10
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    In the previous comment the $-\frac{\ln(b+1)}{2}$ should be $-\frac{\ln(b - 1)}{2}$. Sorry about that. Indeed the domain is $b > 1$. – Dstarred Nov 08 '21 at 21:35