Need calculus help! **Homework Help **

[alfa147x]

I have a couple of problem that look like this:
logb2≈0.29, logb3≈0.46, and logb5≈0.67
(b is some positive real number ≠1)

logb10≈?

"b"= subscripts

If anyone could show me how to do this it would be great!

Thanks,
Alfa147x

 

[alfa147x]

nvm I think we got it

Edit:
Nope... Still need help we get 0.94 :\

 

[DrPizza]

10 is 2*5
Use your properties of logs to break the big log into two baby logs.

edit:
i.e. logAB = logA + logB

 

[alfa147x]

10 is 2*5
Use your properties of logs to break the big log into two baby logs.

edit:
i.e. logAB = logA + logB

Thanks! That was by far the easiest yet we got stuck 😀

 

[TecHNooB]

and here i thought you were solving for the base.

 

[Hacp]

I have a couple of problem that look like this:
logb2≈0.29, logb3≈0.46, and logb5≈0.67
(b is some positive real number ≠1)

logb10≈?

"b"= subscripts

If anyone could show me how to do this it would be great!

Thanks,
Alfa147x

Are you sure this is a calculus problem and not an algebra problem?

I would solve for b and average them out. That should be a decent approximation unless you have b as a function of something else.

Edit: NVM, you have b as subscripts. lol.

Ok this is easy. Just put the numbers into excel, add a trendline and tell it to display an equation. Then use that equation to extrapolate to 10. The value I got for logb10=.957952

 

[alfa147x]

Are you sure this is a calculus problem and not an algebra problem?

I would solve for b and average them out. That should be a decent approximation unless you have b as a function of something else.

Edit: NVM, you have b as subscripts. lol.

lmao it's for a calc class...

 

[DrPizza]

Well, pre-emptive for when you post the next question, typically in homework sets like that, you'll have something like logb150b²

= logb 5²*2*3*b²
=logb5² + logb2 + logb3 + logbb²
=2logb5 + logb2 + logb3 + logbb²
(on that last term, you could move the exponent 2 out in front if you wanted)

Now, a log is an exponent. logb8 means "what's the exponent on b to get 8?"
logbb² means "what's the exponent on b so that you get b²
Well, duh, it's 2.

Or, if you moved the 2 out in front, the question would be
2logbb ---> "2 times 'what's the exponent on b so that you get b?'"
= 2 times 1, so it's still 2.

 

[alfa147x]

Well, pre-emptive for when you post the next question, typically in homework sets like that, you'll have something like logb150b²

= logb 5²*2*3*b²
=logb5² + logb2 + logb3 + logbb²
=2logb5 + logb2 + logb3 + logbb²
(on that last term, you could move the exponent 2 out in front if you wanted)

Now, a log is an exponent. logb8 means "what's the exponent on b to get 8?"
logbb² means "what's the exponent on b so that you get b²
Well, duh, it's 2.

Or, if you moved the 2 out in front, the question would be
2logbb ---> "2 times 'what's the exponent on b so that you get b?'"
= 2 times 1, so it's still 2.

Thanks Doc! It's funny that we started from the back and work our way to the front 😀

 

[DrPizza]

As Hacp suggested, an alternate method would be to solve for b.
Not sure why he was stopped in his tracks because you said it was a subscript.

logb2≈0.29
means 0.29 is the exponent on b to get 2.
or
b^0.29 = 2
Take the log of both sides (natural log or base 10 log, or some other base, if your calculator has that capability. You're in calculus - natural logs are your friends. Screw base 10 logs, they just make problems slightly more cumbersome. Then, use the properties of logs to again solve for b.

That technique would be an incredible waste of time though, since the original problem takes about 3 seconds to do mentally.

 

[DrPizza]

oh, and natural log
ln

means log base e. (Euler's number, pronounced "oiler's")

So
ln(e&#178😉
means "what's the exponent on e so that you get e²"
Again, duhhh, 2.
That's why ln and e^x are inverse functions. It's sort of like squaring a square root.

 

[Hacp]

lmao it's for a calc class...

You don't need any calculus to solve it. If you want to solve it analytically, you just need to know that logb(x)=log10(x)/log10(b). Then solve for b analytically.

 

[DrPizza]

You don't need any calculus to solve it. If you want to solve it analytically, you just need to know that logb(x)=log10(x)/log10(b). Then solve for b analytically.

Very good. You know the change of base formula. Piece of cake. You got that far. I (of course) know how to solve that for b. Go for it - how are YOU going to solve that equation for b?? I'm going to guess that you just painted yourself into a corner.

And, pretty much every calculus book out there has a section on exponential functions and logarithmic functions prior to introducing the derivatives of each. (Seems to be about a month late in the semester for that.)


Also, OP, just rip the log base 10 button off the calculator. Don't bother with it in calculus. If you see a log base 10 of something, use the change of base formula to switch it to natural logs. Natural logs are your friends in calculus. (Unless you want to memorize a few extra formulas.)