Asked by Brooke
Solve the equation : 8^2x-1 = 16^x+1
i know you have to log both sides
log8^2x-1 = log16^x+1
and then you use the exponential rule
2x-1 log8 = x+1 log16
but then i don't know what to do...
i know you have to log both sides
log8^2x-1 = log16^x+1
and then you use the exponential rule
2x-1 log8 = x+1 log16
but then i don't know what to do...
Answers
Answered by
MathMate
Your steps are correct if the equation was:
8^(2x-1) = 16^(x+1)
and assuming this is the case.
Since 8=2^3, and 16=2^4,
you can simplify further using the same rule of logarithms:
log(a^b)=b×log(a)
8^(2x-1) = 16^(x+1)
and assuming this is the case.
Since 8=2^3, and 16=2^4,
you can simplify further using the same rule of logarithms:
log(a^b)=b×log(a)
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.