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secondary 3 | A Maths
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can someone explain why this question need to introduce log but some other questions like e^3x-1=8 dont need
Date Posted: 3 years ago
Views: 190
for the example with e instead, you will have to introduce ln instead, since ln(e)=1. from there, the problem solving for the variable x will be much easier.
you need to introduce log for the question in the picture as you need to bring the unknown variable y down in order the solve it, and you only can do so by adding log.
hope this helps (:
you need to introduce log for the question in the picture as you need to bring the unknown variable y down in order the solve it, and you only can do so by adding log.
hope this helps (:
thx!! :)
Sophie is slightly inaccurate in the first paragraph. There is nothing wrong with applying lg to e, although what Sophie is trying to explain is that e is better paired with ln and 10 is better paired with lg as part of simplifications.
e^(2x + 3) = 6
We can apply ln directly.
ln e^(2x + 3) = ln 6
(2x + 3) ln e = ln 6
(2x + 3) = ln 6 / ln e
2x = (ln 6 / ln 3) - 3
x = 0.5 [(ln 6 / ln 3) - 3]
But ln e = 1, so one of the above steps can be simplified without having to go the long way.
2x + 3 = ln 6
2x = (ln 6) - 3
x = 0.5 [(ln 6) - 3]
Or, we can apply lg directly.
lg e^(2x + 3) = lg 6
(2x + 3) lg e = lg 6
But the value lg e is not easy to predict.
2x + 3 = (lg 6 / lg e)
2x = (lg 6 / lg e) - 3
x = 0.5 [(lg 6 / lg 3) - 3]
LockB, applying lg and ln works for every such exponential equation, regardless of whether the base is 10, e or some other positive number not equalling 1.
If calculators are made to calculate logs to base 2 as well, then changing to base 2 would even perhaps be possible.
e^(2x + 3) = 6
We can apply ln directly.
ln e^(2x + 3) = ln 6
(2x + 3) ln e = ln 6
(2x + 3) = ln 6 / ln e
2x = (ln 6 / ln 3) - 3
x = 0.5 [(ln 6 / ln 3) - 3]
But ln e = 1, so one of the above steps can be simplified without having to go the long way.
2x + 3 = ln 6
2x = (ln 6) - 3
x = 0.5 [(ln 6) - 3]
Or, we can apply lg directly.
lg e^(2x + 3) = lg 6
(2x + 3) lg e = lg 6
But the value lg e is not easy to predict.
2x + 3 = (lg 6 / lg e)
2x = (lg 6 / lg e) - 3
x = 0.5 [(lg 6 / lg 3) - 3]
LockB, applying lg and ln works for every such exponential equation, regardless of whether the base is 10, e or some other positive number not equalling 1.
If calculators are made to calculate logs to base 2 as well, then changing to base 2 would even perhaps be possible.
10^(1 - y) = 14
lg is a good idea here.
lg 10^(1 - y) = lg 14
(1 - y) lg 10 = lg 14
But we know that lg 10 = 1, so...
1 - y = lg 14
y = 1 - lg 14
ln is a good idea as well.
ln 10^(1 - y) = ln 14
(1 - y) ln 10 = ln 14
Sad that ln 10 has no easy value, just like lg e.
(1 - y) = (ln 14 / ln 10)
y = 1 - (ln 14 / ln 10)
For questions with bases other than 10 and e, there is practically little difference in the workings other than the logarithm applied.
lg is a good idea here.
lg 10^(1 - y) = lg 14
(1 - y) lg 10 = lg 14
But we know that lg 10 = 1, so...
1 - y = lg 14
y = 1 - lg 14
ln is a good idea as well.
ln 10^(1 - y) = ln 14
(1 - y) ln 10 = ln 14
Sad that ln 10 has no easy value, just like lg e.
(1 - y) = (ln 14 / ln 10)
y = 1 - (ln 14 / ln 10)
For questions with bases other than 10 and e, there is practically little difference in the workings other than the logarithm applied.
You will be able to calculate the logs with other bases with a Graphing Calculator (GC) or computer software. The GC is used in the syllabi of the Singapore A level Mathematics subjects.
But at O level, you'll need to change base since the approved scientific calculators do not have that function.
But at O level, you'll need to change base since the approved scientific calculators do not have that function.
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An alternative approach.
Here, both lg and ln work perfectly fine! Ln can also be used here.
For e^(3x - 1) = 8, there is nothing wrong with using lg to solve it, while for 10^(3x - 1) = 8, there is nothing wrong with using ln to solve it (if calculators can calculate logs to other bases also, it would also be allowed).
But e is commonly paired with ln because ln e = 1 as part of the workings and this simplifies our calculations a little.
Which is why ln is commonly applied to e and lg is commonly applied to 10.
Here, both lg and ln work perfectly fine! Ln can also be used here.
For e^(3x - 1) = 8, there is nothing wrong with using lg to solve it, while for 10^(3x - 1) = 8, there is nothing wrong with using ln to solve it (if calculators can calculate logs to other bases also, it would also be allowed).
But e is commonly paired with ln because ln e = 1 as part of the workings and this simplifies our calculations a little.
Which is why ln is commonly applied to e and lg is commonly applied to 10.
Date Posted:
3 years ago
thx :)
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