Logarithm help as my previous post
- Thread starter 8dltndus
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Did you notice that 4 = 22? Use the change-of-base formula, like with the other equation you posted:
. . . . .\(\displaystyle \log_4(8y\, +\, 1)\, =\, \dfrac{\log_2(8y\, +\, 1)}{\log_2(4)}\, =\, \dfrac{\log_2(8y\, +\, 1)}{2}\)
...and so forth.
HallsofIvy
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In this particular case, because \(\displaystyle 4= 2^2\) you don't really need to use the full "change of base" formula.
If \(\displaystyle x= log_4(8y+ 1)\) then \(\displaystyle 8y+ 1= 4^x= (2^2)^x= (2^x)^2\) so that \(\displaystyle 2^x= \sqrt{8y+ 1}\) and so \(\displaystyle x= log_2(\sqrt{8y+ 1})\). So your equation becomes \(\displaystyle log_2(\sqrt{8y+1})= log_2(3y)\) and then \(\displaystyle \sqrt{8y+ 1}= 3y\). That is, of course, just what you get from stapel's suggestion.
If \(\displaystyle x= log_4(8y+ 1)\) then \(\displaystyle 8y+ 1= 4^x= (2^2)^x= (2^x)^2\) so that \(\displaystyle 2^x= \sqrt{8y+ 1}\) and so \(\displaystyle x= log_2(\sqrt{8y+ 1})\). So your equation becomes \(\displaystyle log_2(\sqrt{8y+1})= log_2(3y)\) and then \(\displaystyle \sqrt{8y+ 1}= 3y\). That is, of course, just what you get from stapel's suggestion.