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The default equation is $(1 + x)^3=4^{-y^2} $

I solved as follows:

$(-3\log_4(1+x))^{1/2}=y$

With the logarithm base equal to $4$, my idea is that $4$ is the number we have in the right part of the equation, that's why I went with $4$;

And I assume this is the right answer as well, but the actual book's answer says:

$3 \ln(1+x)=-y^2 \ln(4)$;

$\left(-\frac{3}{\ln(4)} \ln(1+x)\right)^{1/2}=y$

How did they went with logarithm base equal to $\mathrm{e}$? Thus, I assume you could insert any base there. I don't understand this step. Could you please explain this case of bringing down the equation to the same logarithmic base, and any base at all as it seems. Or provide me with resources for me to read there. Thank you!!!

Googled for advanced logarithmic equations, nothing relevant shows up.

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  • $\begingroup$ You should know $\log_a(b)\times\log_b(c)=\log_a(c)$. From that you can deduce everything. $\endgroup$ – user202729 Apr 18 '16 at 8:32
  • $\begingroup$ No need to shout. $\endgroup$ – Yves Daoust Apr 18 '16 at 8:51
  • $\begingroup$ Resources to read: any high school book that introduces logarithms. $\endgroup$ – fqq Apr 18 '16 at 8:55
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For any base $b$,

$$\log_b(x)=\frac{\ln(x)}{\ln(b)}.$$


This is because

$$x=b^{\log_b(x)}=(e^{\ln(b)})^{\log_b(x)}=e^{\ln(b)\log_b(x)}=e^{\ln(x)}.$$

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