+1124 Hi friends,
I promise..only 2 more questions....
I have a basic understanding of natural logs, the expression being ln, which is the same as log with base e...and so on...I am just not sure how to handle this one:
Solve the following using natural logs:
A=1.152∗√5.46.21
All and any help is always greatly appreciated!
+118702 I really do not know what you are expected to do here.
Maybe
A=1.152∗√5.46.21lnA=ln1.152+ln5.40.5−ln6.21lnA=2ln1.15+0.5ln5.4−ln6.21lnA≈−0.7034..A≈0.49488
Why anyone would bother with that I have no idea.
The exact answer, without logs, is 23√15180
LaTex
A={{1.15^2* \sqrt{5.4}} \over6.21}\\
lnA=ln1.15^2+ln5.4^{0.5}-ln6.21\\
lnA=2ln1.15+0.5ln5.4-ln6.21\\
lnA\approx -0.7034..\\
A\approx 0.49488
+1124 hi Melody,
yes, it was copied correctly....Thank you for the answer....
