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Help with natural logarithms?

User947387

I thought that 5x+1=ln20 therefore, 5x=ln 19 x= ln/5 (wrong!) :frown:

(edited 8 years ago)

Reply 1

User1686961

Original post by apronedsamurai

e^5x+1=20

I thought that 5x+1=ln20 therefore, 5x=ln 19 x= ln/5 (wrong!) :frown:

is that E^(5x+1) or E^(5x)+1 ?

Reply 2

User947387

Original post by ThatGuyRik

is that E^(5x+1) or E^(5x)+1 ?

Tidied up the notation :smile:

Reply 3

User1686961

Original post by apronedsamurai

Tidied up the notation :smile:

take ln of both sides. you get ln20 and 5x+1.

You should be able to take it from here.

Reply 4

User947387

Original post by ThatGuyRik

take ln of both sides. you get ln20 and 5x+1.

You should be able to take it from here.

And when I do, I get; 5x+1=ln20 therefore, 5x=ln 19 x= ln/5 which is wrong

Reply 5

MichuAtDeGeaBa

e^(5x+1) = 20

5x+1 = Ln20

5x = (ln20) - 1

x = ((ln20)-1)/5)

x = 0.399

Reply 6

User1686961

Original post by apronedsamurai

And when I do, I get; 5x+1=ln20 therefore, 5x=ln 19 x= ln/5 which is wrong

ln20 is in its own form. So if you take -1 from both sides you get 5x=ln(20)-1 not ln(19).

Reply 7

User1686961

Original post by MichuAtDeGeaBa

e^(5x+1) = 20

5x+1 = Ln20

5x = (ln20) - 1

x = ((ln20)-1)/5)

x = 0.399

You're not meant to give them the answer. Meant to guide them to it.

Reply 8

MichuAtDeGeaBa

Original post by ThatGuyRik

You're not meant to give them the answer. Meant to guide them to it.

It's a tiny mistake that Ln20 - 1 is that and not Ln19, not a procedural error

You've pointed it out your self. I'm sure if the person is studying logarithm's then he/she will be able to rearrange simple equations

Reply 9

User1686961

Original post by MichuAtDeGeaBa

Well still I've been told not to give a direct answer. Show them the mistake instead.

Reply 10

Notnek

Original post by MichuAtDeGeaBa

It's a tiny mistake that Ln20 - 1 is that and not Ln19, not a procedural error

So why not just say this instead of giving a full solution?

Whatever your opinions, full solutions are against the rules in this forum so please try to only use them as a last resort :smile:

Reply 11

User947387

Original post by ThatGuyRik

ln20 is in its own form. So if you take -1 from both sides you get 5x=ln(20)-1 not ln(19).

But why is that the case?

Reply 12

User1686961

Original post by apronedsamurai

But why is that the case?

Taking ln of E^(5x+1) gives 5X+1. the otherside is ln(20).

ln(1)=0 So either way you cant do ln(20)-1=ln(19).

Basics tell you this anyway so you should be able to figure that out.

Reply 13

davros

Original post by apronedsamurai

But why is that the case?

Because of the way logarithms work.

It's not true that

log a - b = log(a - b)

You have a log law which says that

log(a/b) = log a - log b

so in your case you would have

(ln20) - 1 = (ln 20) - (ln e) = ln(20/e), not ln(19)

Reply 14

User947387

Original post by davros

So even with the natural logarithms, the three basic rules of logarithms, or the "log rules" will ALWAYS be applied, and ALWAYS relevant?

Reply 15

davros

Original post by apronedsamurai

So even with the natural logarithms, the three basic rules of logarithms, or the "log rules" will ALWAYS be applied, and ALWAYS relevant?

The "log rules" apply to logarithms of all bases, and in particular natural logarithms, which are just "logs to base e" :smile:

Reply 16

User947387

Original post by davros

The "log rules" apply to logarithms of all bases, and in particular natural logarithms, which are just "logs to base e" :smile:

log(a/b) = log a - log b: is that rule commutative?

Reply 17

davros

Original post by apronedsamurai

log(a/b) = log a - log b: is that rule commutative?

Not sure what you mean.

"Rules" aren't commutative; operations are. So the operation of addition is commutative because a + b = b + a.

If you define an operation $ by a$b = log a - log b, then b$a = log b - log a = -(log a - log b) which is not the same thing as a$b. So that operation that I've called '$' is not commutative.