Question 248310
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I presume you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4(2)\ +\ \log_4(x)\ =\ \log_4(26)]


Which, if properly said in normal language would have been "log <i>to the base 4 of 2</i> plus ...and so on.


The sum of the logs is the log of the product, which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) + \log_b(y) = \log_b(xy)]


Or for your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4(2)\ +\ \log_4(x)\ =\ \log_4(2x)]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4(2x)\ =\ \log_4(26)]


If equal base logs are equal, then their arguments must be equal, which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x)\ =\ \log_b(y) \ \Leftrightarrow\ \ x\ =\ y]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ 2x\ =\ 26]


You can take it from there.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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