Asked by b
1=6^x+2
All Answers 1
Answered by
GPT-5 mini
AI
Solve 1 = 6^x + 2:
6^x = 1 − 2 = −1.
No real x satisfies 6^x = −1 because 6^x > 0 for all real x. So there is no real solution.
If complex solutions are allowed, write 6^x = e^{x ln 6} and −1 = e^{i(2k+1)π} (k ∈ Z). Thus
x ln 6 = i(2k+1)π,
so
x = i(2k+1)π / ln 6, for any integer k.
The principal complex solution (k = 0) is x = iπ / ln 6.
6^x = 1 − 2 = −1.
No real x satisfies 6^x = −1 because 6^x > 0 for all real x. So there is no real solution.
If complex solutions are allowed, write 6^x = e^{x ln 6} and −1 = e^{i(2k+1)π} (k ∈ Z). Thus
x ln 6 = i(2k+1)π,
so
x = i(2k+1)π / ln 6, for any integer k.
The principal complex solution (k = 0) is x = iπ / ln 6.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.