Solve ln (y - 2) = t - 5 for y?

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Solve ln (y - 2) = t - 5 for y?

2007-09-10 17:08:11 · 4 answers · asked by jewel7962002 1 in Science & Mathematics ➔ Mathematics

4 answers

Lets do this one right, shall we????

The above answers are terribly wrong.....as of this posting......9:28:28 pm PST

take the exponential log base , e, of both sides...
e^( ln ( y - 2 )) = e^ ( t - 5 )

y - 2 = e^ ( t - 5 )

y = e^ ( t - 5 ) + 2

by the way.......e^(ln A ) = A, and ln( e^B) = AB..... properties of logs and exponentials.......

2007-09-10 17:22:12 · answer #1 · answered by Mathguy 5 · 0⤊ 1⤋

y=t-3

2007-09-11 00:19:23 · answer #2 · answered by Anonymous · 0⤊ 0⤋

y-2=t-5
y=t-5+2=t-3 ans

2007-09-11 00:18:28 · answer #3 · answered by alpha 7 · 0⤊ 0⤋

e^(t - 5) = y - 2
y = 2 + e^(t - 5)

2007-09-11 07:11:02 · answer #4 · answered by Como 7 · 1⤊ 0⤋