SOLUTION: Log x^5=27, x=.......?
first u must plug the Logarithm back into exponential form... which is b^x=y (if u dont know the equation for the log it is log b^y=x)
that would then ma
Algebra ->
Logarithm Solvers, Trainers and Word Problems
-> SOLUTION: Log x^5=27, x=.......?
first u must plug the Logarithm back into exponential form... which is b^x=y (if u dont know the equation for the log it is log b^y=x)
that would then ma
Log On
Question 195267: Log x^5=27, x=.......?
first u must plug the Logarithm back into exponential form... which is b^x=y (if u dont know the equation for the log it is log b^y=x)
that would then make the equation X^27=5
the you would take the 27 root of 5 and that with give you x
which should equal 1.06142125 which rounded is 1.1
You can put this solution on YOUR website! first u must plug the Logarithm back into exponential form... which is b^x=y (if u dont know the equation for the log it is log b^y=x)
that would then make the equation X^27=5
the you would take the 27 root of 5 and that with give you x
which should equal 1.06142125 which rounded is 1.1
(