Just wondering if someone could show me the working out for this question?
Thanks
2007-06-27 22:21:54 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
as u have used the term Ln ,i would assume the base as 'e'...had it been log then one should have used 10 as base....nyhow
ln(3x-4)=1.6
3x-4=e^1.6
3x-4=4.95.....(calculate using calci)
3x=8.95
x=2.98
hope this would have helped
2007-06-27 22:28:37 · answer #1 · answered by razmatttaz... 4 · 0⤊ 0⤋
(3x - 4) = e^(1.6)
3x - 4 = 4.95
3x = 8.95
x = 2.983
Check
ln (3 x 2.983 - 4) = 1.6 as required.
2007-07-02 03:05:26 · answer #2 · answered by Como 7 · 0⤊ 0⤋
ln(3x-4) = 1â6 (take anti-ln).
3x - 4 = 4â953 032 424...
3x = 4â953 032 424... + 4
3x = 8â953 032 424...
x = 8â953 032 424.../ 3
x = 2â984 344 141...
x â 2â984
2007-06-28 06:31:30 · answer #3 · answered by Sparks 6 · 0⤊ 0⤋
e^[ln(3x-4)] = e^(1.6) ...... taking antilog of natural log on both sides...
which will give,
3x-4 = e^(1.6)
3x = e^(1.6) + 4
x = [e^(1.6) + 4] / 3
2007-06-28 05:29:12 · answer #4 · answered by Sindhoor 2 · 0⤊ 0⤋
ln(3x-4=1.6
3x-4=e^(1.6)
3x=4+e^(1.6)=4+4.95=8.95
x=2.983. answer
2007-06-28 06:01:05 · answer #5 · answered by Anonymous · 0⤊ 0⤋
ln(3x-4)=1.6
3x-4 = e^1.6
3x-4 = 2.718^1.6
3x-4 = 4.953
3x = 8.953
x = 8.953/3
x = 2.984...........
2007-06-28 05:28:14 · answer #6 · answered by harry m 6 · 0⤊ 0⤋