Asked by kayla
how do i solvethis equation (btw we are studyinglogarithms)
3^(x^3)=9^x
2^x=10
(4^x)-(2^x)=0
3^(x^3)=9^x
2^x=10
(4^x)-(2^x)=0
Answers
Answered by
Reiny
for the first you don't even need logs
3^(x^3)=9^x
3^(x^3)=(3^2)^x
3^(x^3) = 3^(2x)
so x^3 = 2x
x^3 - 2x = 0
x(x^2 - 2) = 0
x = 0 or x = ±√2
for second, take logs of both sides
log(2^x) = log 10
xlog2 = 1
x = 1/log2
for the last
4^x = 2^x
2^(2x) = 2^x
2x = x
x = 0
3^(x^3)=9^x
3^(x^3)=(3^2)^x
3^(x^3) = 3^(2x)
so x^3 = 2x
x^3 - 2x = 0
x(x^2 - 2) = 0
x = 0 or x = ±√2
for second, take logs of both sides
log(2^x) = log 10
xlog2 = 1
x = 1/log2
for the last
4^x = 2^x
2^(2x) = 2^x
2x = x
x = 0
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.