How do you solve #ln y = 2t - 3#?

1 Answer
mason m
Jan 28, 2016

#y=e^(2t-3),t=(lny+3)/2#

Explanation:

Solving for #mathbfy#:

To undo the natural logarithm that #y# is attached to, exponentiate both sides with base #e#. Since exponentiation and logarithms are inverse functions, and #ln# is equivalent to #log_e#, these will undo one another and leave us with just #y#.

#e^lny=e^(2t-3)#

#y=e^(2t-3)#

Solving for #mathbft#:

This is simpler than solving for #y#. Add the #3# and divide both sides by #2# to see that

#2t=lny+3#

#t=(lny+3)/2#

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