Question 249764
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*[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]
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