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Solve for x using logarithms

Devi

If anybody could explain this it would be a real help.

5^(-x+3)=1

This is not the actual question I have a problem with on my assignment, I just want to learn how to solve them.

Cheers Devi

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Pole Monkey

Take the log of both sides and use the identity,

[latex]log{a^b} = bloga[/latex]

Yakuza

[LATEX]
5^{(-x+3)}=1
[/LATEX]

To solve this using logs, the thing I would do is to take logs of both sides (to the base of the number you're raising the power to).

So the equation now becomes:
-x+3 = log 1 (base 5).

How do we calulate logs in bases other than those in our calculators?
Use the result log a (base b) = ln |a| / ln |b| (ln =natural log) (it can in fact be any base c - log a (base c) / log b (base c), I chose ln as it's on the calculator)

So, what is log 1 in base 5? ln |1| / ln |5| = 0/5 = 0

Therefore: -x+3 = 0, x=3.

In this particular case, you could note that anything^0 = 1, so therefore -x+3 = 0 or x = 3 without having to use logs at all!

If the problem had been [LATEX]5^{(-x+3)}=100[/LATEX], what would your answer be?

Devi

Thanks Yakuza, your post is very clear and I think im getting there just bear with me.
.
In you question $mathtran?tex=5%5E%7B(-x%2B3)%7D%3D100$

(-x+3) = log100(base 5)
-x+3 = ln(100)/ln(5)
-x+3 = 4.60/1.60
-x+3 = 2.87
3-2.87 = x
-0.13 = x

Is that right?

Yakuza

Devi wrote: »

Thanks Yakuza, your post is very clear and I think im getting there just bear with me.
.
In you question $mathtran?tex=5%5E%7B(-x%2B3)%7D%3D100$

(-x+3) = log100(base 5)
-x+3 = ln(100)/ln(5)
-x+3 = 4.60/1.60
-x+3 = 2.87
3-2.87 = x
-0.13 = x

Is that right?

Almost - take a look at the second last and last lines of your work again, watching for signs.

Devi

Yakuza wrote: »

Almost - take a look at the second last and last lines of your work again, watching for signs.

Should I have moved the 3 to the other side instead?

-x+3 = 2.87
-x = 2.87-3
-x = -0.13
x = 0.13

Yakuza

Your only mistake was in moving from second last line to the last line:

3-2.87 = x
-0.13 = x

should have read:

3-2.87 = x
0.13 = x

Taking 2.87 from 3 shouldn't give a negative number!

Other than that, the rest was fine in principle - I'd give an answer to 4 or 5 decimal places when dealing with logs, though.
5^2.87 = 101.4014
5^2.8614 = 100.0075

Devi

Yakuza wrote: »

Your only mistake was in moving from second last line to the last line:

should have read:

Taking 2.87 from 3 shouldn't give a negative number!

Other than that, the rest was fine in principle - I'd give an answer to 4 or 5 decimal places when dealing with logs, though.
5^2.87 = 101.4014
5^2.8614 = 100.0075

Oh right, silly mistake. Looking at this stuff all day, brain is starting to get tired. They only ask for two in the assignment so thats why i did two.

Thanks Yakuza youve been a great help. Im sure this isnt the last youll here from me lol.

Yakuza

You're very welcome, glad to have helped.

Just one last thing, without wanting to melt your head. While the assignment might have asked you for an answer to 2 dp, that's fair enough, it's always good to use more dp for your interim calculations, so you don't lose accuracy.

In doing what you did (expressing ln 100 as 4.60 and ln 5 as 1.60 and dividing the two), you end up with 2.87 and an answer of 0.13, whereas the "true" answer would be closer to 0.14. That's over 7% of a difference - which can be significant enough depending on the field to which the problem applied. If an actuary calculated a premium that was 7% too small on a number of cases, their employer wouldn't be too happy with them, or if they calculated one that was 7% too high, the customer mightn't like it and go elsewhere etc.

By using 4dp for the interim steps so getting 4.6052 and 1.6094 you end up with 2.8614 and an answer of 0.1386 or 0.14 to 2 dp

Devi

Yakuza wrote: »

You're very welcome, glad to have helped.

Just one last thing, without wanting to melt your head. While the assignment might have asked you for an answer to 2 dp, that's fair enough, it's always good to use more dp for your interim calculations, so you don't lose accuracy.

In doing what you did (expressing ln 100 as 4.60 and ln 5 as 1.60 and dividing the two), you end up with 2.87 and an answer of 0.13, whereas the "true" answer would be closer to 0.14. That's over 7% of a difference - which can be significant enough depending on the field to which the problem applied. If an actuary calculated a premium that was 7% too small on a number of cases, their employer wouldn't be too happy with them, or if they calculated one that was 7% too high, the customer mightn't like it and go elsewhere etc.

By using 4dp for the interim steps so getting 4.6052 and 1.6094 you end up with 2.8614 and an answer of 0.1386 or 0.14 to 2 dp

Right I know what you mean, better go over my assignment again to check that. Thanks again.

FoxT

Yakuza wrote: »

[LATEX]
5^{(-x+3)}=1
[/LATEX]

To solve this using logs, the thing I would do is to take logs of both sides (to the base of the number you're raising the power to).

So the equation now becomes:
-x+3 = log 1 (base 5).

How do we calulate logs in bases other than those in our calculators?
Use the result log a (base b) = ln |a| / ln |b| (ln =natural log) (it can in fact be any base c - log a (base c) / log b (base c), I chose ln as it's on the calculator)

So, what is log 1 in base 5? ln |1| / ln |5| = 0/5 = 0

Therefore: -x+3 = 0, x=3.

In this particular case, you could note that anything^0 = 1, so therefore -x+3 = 0 or x = 3 without having to use logs at all!

If the problem had been [LATEX]5^{(-x+3)}=100[/LATEX], what would your answer be?

I have a question.... how does using Base 5 help?

5^(-x+3) = 100

(-x+3)*Log(5) = Log 100 /*note: the base has no effect. though ofc it must be the same base on each side */

(-x+3) = Log(100)/Log(5)

.
.
.

did I miss something?

Thx

Yakuza

You're quite right, it's just that I personally find it easier to solve problems like this by getting rid of the exponent (taking the log to the base of the number that the unknown is raised to). I did say it was the way I would do it, your method (or the one hinted at by Pole Monkey) are perfectly fine. There isn't a huge difference between what we do in any case. I haven't formally studied logs/powers in close to 30 years, so some of their rules don't come to mind as quickly as others, so "I goes with what I knows"