Logarithmic Equation

[nycmathguy]

Find x.

 

[MathLover1]

log(x+1)=log(x)+1

log(x+1)-log(x)=1

log((x+1)/x)=1..........1=log(10)

log((x+1)/x)=log(10)

(x+1)/x=10

x+1=10x

10x-x=1

9x=1

x = 1/9

 

[nycmathguy]

log(x+1)=log(x)+1

log(x+1)-log(x)=1

log((x+1)/x)=1..........1=log(10)

log((x+1)/x)=log(10)

(x+1)/x=10

x+1=10x

10x-x=1

9x=1

x = 1/9

Good to review this stuff.

 

[Country Boy]

That is assuming the logarithm is the "common logarithm", base 10. If it is the "natural logarithm", base e, or a logarithm to some other base, use the same method but you will get a different answer.

For example, if the logarithm is the "natural logarithm then log(e)= 1 so the equation becomes log(x+ 1)= log(x)+ log(e). log((x+ 1)/x)= log(e) and then (x+1)/x= e. x+ 1= ex. (e- 1)x= 1. x= 1/(e- 1).

More generally, for any base, a, the equation log(x+1)= log(x)+ 1 has solution 1/(a- 1).

 

[MathLover1]

That is assuming the logarithm is the "common logarithm", base 10. If it is the "natural logarithm", base e, or a logarithm to some other base, use the same method but you will get a different answer.

For example, if the logarithm is the "natural logarithm then log(e)= 1 so the equation becomes log(x+ 1)= log(x)+ log(e). log((x+ 1)/x)= log(e) and then (x+1)/x= e. x+ 1= ex. (e- 1)x= 1. x= 1/(e- 1).

More generally, for any base, a, the equation log(x+1)= log(x)+ 1 has solution 1/(a- 1).

yes, base 10 is always assumed if you write log(x)
if you write ln(x) means you are dealing with natural logarithm

 

[Country Boy]

In secondary math, yes. In university math, which uses the natural logarithm almost exclusively, log(x) is typically assumed to be the natural logarithm unless otherwise specified.

 

[MathLover1]

In secondary math, yes. In university math, which uses the natural logarithm almost exclusively, log(x) is typically assumed to be the natural logarithm unless otherwise specified.

In college, it is not true that log(x)=log(e,x). Rarely is it used that way. ln is by far more pervasive in undergrad and grad students, meaning log(e,x). In fact, any undergraduate student overlapping computer science and math will suggest that log should be interpreted to mean log(2,x).
It is possible that in a few nations, the preference at the undergraduate level deviate from this, but if that's the case, you should understand that doing so is an exception, not the rule.

 

[nycmathguy]

yes, base 10 is always assumed if you write log(x)
if you write ln(x) means you are dealing with natural logarithm

Really? I didn't know this fact.

 

[nycmathguy]

That is assuming the logarithm is the "common logarithm", base 10. If it is the "natural logarithm", base e, or a logarithm to some other base, use the same method but you will get a different answer.

For example, if the logarithm is the "natural logarithm then log(e)= 1 so the equation becomes log(x+ 1)= log(x)+ log(e). log((x+ 1)/x)= log(e) and then (x+1)/x= e. x+ 1= ex. (e- 1)x= 1. x= 1/(e- 1).

More generally, for any base, a, the equation log(x+1)= log(x)+ 1 has solution 1/(a- 1).

Thanks. I know you from another math site but just can't remember the name.

 

[nycmathguy]

In college, it is not true that log(x)=log(e,x). Rarely is it used that way. ln is by far more pervasive in undergrad and grad students, meaning log(e,x). In fact, any undergraduate student overlapping computer science and math will suggest that log should be interpreted to mean log(2,x).
It is possible that in a few nations, the preference at the undergraduate level deviate from this, but if that's the case, you should understand that doing so is an exception, not the rule.

Are you saying that math, the universal language of numbers, is calculated differently in nations outside the United States?

 

[MathLover1]

Are you saying that math, the universal language of numbers, is calculated differently in nations outside the United States?

s
no, nothing is calculated differently outside the United States, there are just some differences regarding the "common logarithm", base 10,
when I was a student, we were assuming the log(x) is log(10, x)

 

[nycmathguy]

s
no, nothing is calculated differently outside the United States, there are just some differences regarding the "common logarithm", base 10,
when I was a student, we were assuming the log(x) is log(10, x)

Ok. Posting trigonometric equations in a few minutes.