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Show $\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$

from LS

$\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(3/2) + \log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2\big(\frac{6(n+1)}{6}\big)}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$

is this right?

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Yes, your solution is correct, but you could have done it a more "natural" way. In this case of your problem, there two useful properties of logarithms that you could use: exponents can be brought out front and multiplication under the logarithm sign turns into addition when the multiplicands are separated.

1)$$ \log_{a}{\frac{x}{y}}=\log_{a}{\left[\left(\frac{y}{x}\right)^{-1}\right]}= -1\cdot\log_{a}{\frac{y}{x}}=-\log_{a}{\frac{y}{x}}. $$

2)$$ \log_{a}\left({xy}\right)=\log_{a}{x}+\log_{y}. $$


$$ 1+\frac{\log_2\left[\frac{2}{3}(n+1)\right]}{\log_2\frac{3}{2}} = 1+\frac{\log_2\frac{2}{3}+\log_2\left(n+1\right)}{\log_2\frac{3}{2}} =\\ 1+\frac{-\log_2\frac{3}{2}+\log_2\left(n+1\right)}{\log_2\frac{3}{2}}= 1-\frac{\log_2\frac{3}{2}}{\log_2\frac{3}{2}}+\frac{\log_2\left(n+1\right)}{\log_2\frac{3}{2}}=\\ 1-1+\frac{\log_2\left(n+1\right)}{\log_2\frac{3}{2}}= \frac{\log_2\left(n+1\right)}{\log_2\frac{3}{2}} $$

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Its right. Although for clarity, perhaps it would be better to write $\displaystyle \frac{\log_2(2/3(n+1))}{\log_2(3/2)}$ as $\displaystyle \frac{\log_2((2/3)(n+1))}{\log_2(3/2)}$

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I would write $$\log_{2}\frac{3}{2}+\log_{2}\frac{2}{3}+\log_{2}(n+1)=\log_{2}(n+1)$$ since $$\log_{2}{\frac{2}{3}}=-\log_{2}{\frac{3}{2}}$$

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