solve exponential logarithmic equations calculator

This equation has a strictly-numerical term.
No, the two values are not equal, but they're pretty darned close.
Solve log2( x ).5, accurate to two decimal places.
Sometimes, a particular solution won't work.
Then I'll set the arguments equal.If, on the other hand, my solution had returned a value of, say,.083, then I would have known that, no, my answer was wrong.On the left-hand side, I have only just the one log backup my windows 8.1 term.For instance, to check the solution of the equation log2(3 x ).5, I'll plug.54 in for x, and see how close the result is.5: log2(3 x ) log2(3(7.54) log2(22.62) At this point, I'll need to use the change-of-base formula to convert.Don't blindly assume that every negative solution must be disallowed.

I'll use the natural log: log2(22.62) ln(22.62) ln(2).
This equation has a strictly numerical term (being the 3 on the right-hand side).
But whereas something like 23 can be simplified to a straight-forward 8, the irrational value of e 3 can only be approximated in the calculator.
So: log3(3)2 6(3) log3(9 18) log3(27) log3(33) 3 Huh.Try to further simplify, related, graph mathrm Plotting: Sorry, your browser does not support this application.Dividing both sides by log 3: ( y dfraclog_5log_3 using a calculator we can find that log.69897 and log.4771 2 then our equation becomes: ( n dfraclog_5log_3 dfrac0.698970.47712.46497 ).Solve log3( x 2 6 x ) 3 This log equation has a strictly numerical term, being the 3 on the right-hand side.Allowing for round-off error, these values confirm to me that I've gotten the right answer.So I plug the expression into my calculator, and round the result on my screen.Note that this decimal form is not "better" than e 3; actually, e 3 is the exact, and therefore the more correct, answer.An example would be: Solve ln( x ) 3, giving your answer accurate to three decimal places.Log3( x 2 6 x ) 3 33 x 2 6 x 27 x 2 6 x 0 x 2 6 x 27 0 ( x 9 x 3) x 3, 9 Before I say that x 3, 9 is my solution, I need.However, in this case (maybe leading up to graphing or word problems) they want me to provide a decimal approximation.