Asked by Anonymous
How do I solve 3log10of(x-6)=11 and e^(2x)+10e^(x)-25=0?
Answers
Answered by
Reiny
3log10of(x-6)=11
log 10 (x-6)^3 = 11
(x-6)^3 = 10^11
x-6 = 10^(11/3)
x = 10^(11/3) + 6
for the second, let e^x = a
then we have
a^2 + 10a - 25 = 0
(a-5)^2 = 0
a = +- 5
e^x = 5 or e^x = -5 (the second has no solution)
x = ln5
log 10 (x-6)^3 = 11
(x-6)^3 = 10^11
x-6 = 10^(11/3)
x = 10^(11/3) + 6
for the second, let e^x = a
then we have
a^2 + 10a - 25 = 0
(a-5)^2 = 0
a = +- 5
e^x = 5 or e^x = -5 (the second has no solution)
x = ln5
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.