Question 249764
<font face="Garamond" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6^{2x}\ -\ 42\,\cdot\,6^x\ +\ 216\ =\ 0]


Note that *[tex \LARGE 6^{2x}\ =\ 6^x^2\ =\ 6^x\,\cdot\,6^x]


Which means that the equation takes on the form of a quadratic with the function *[tex \LARGE 6^x] assuming the role of the independent variable.


Let *[tex \Large u] represent *[tex \Large 6^x], then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 42u\ +\ 216\ =\ 0]


Note that *[tex \Large -6\ +\ (-36)\ =\ -42] and  *[tex \Large -6\,\cdot\,-36\ =\ 216] so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (u\ -\ 6)(u\ -\ 36)\ =\ 0]


Replacing *[tex \Large u\ =\ 6^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (6^x\ -\ 6)(6^x\ -\ 36)\ =\ 0]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6^x\ -\ 6\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6^x\ =\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_6\left(6^x\right)\ =\ \log_6(6)]


Recalling:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(b)\ =\ 1]


We can write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\log_6\left(6\right)\ =\ \log_6(6)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 1]


Similarly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6^x\ -\ 36\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6^x\ =\ 36]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_6\left(6^x\right)\ =\ \log_6(36)]


And noting that *[tex \Large 36\ =\ 6^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\log_6\left(6\right)\ =\ 2\log_6(6)]


and finally:


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


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