Graphing log function

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Elliawt

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I simply do not understand the question.. thanks for your help!

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Elliawt said:
I simply do not understand the question.. thanks for your help!

View attachment 29324
Click to expand...
If I were to solve this problem, for my assignment, I would start with:

y = ln (b*x) → b * x = e^ x & y = 0 when x = 2 → b = (1/2) * e^2

continue......

Please show us what you have tried and exactly where you are stuck.

What is the definition of "focal date"?

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.

Oh I would also review my post to make sure that the sketch was posted right-side-up.

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Yes.. I know how to convert log to exp and vice versa but I don't understand the (BX) and what value do I plug in " BSE " ?

f (x) = a log base (bx)... if a = 2;

y = 2 log base (bx)

inverse:

x = 2 log base (by)

x/2 = log base ( by )

base"x/2 = by

f -1 (x) = base"x/2 (by) .... is this correct so far ?
 
Elliawt said:
Yes.. I know how to convert log to exp and vice versa but I don't understand the (BX) and what value do I plug in " BSE " ?

f (x) = a log base (bx)... if a = 2;

y = 2 log base (bx)

inverse:

x = 2 log base (by)

x/2 = log base ( by )

base"x/2 = by

f -1 (x) = base"x/2 (by) .... is this correct so far ?
Click to expand...
I'm baffled. It says a=1; why are you setting it to 2? I'm also curious why they call the base of the logarithm "base" (or "bse"?) rather than using a proper variable?

As I understand the problem, they are defining a function [imath]f(x) = \log_c(bx)[/imath], and telling you that [imath]f(2) = 0[/imath] and [imath]f(3) = 1[/imath]. Plug those two pairs in to obtain two equations in variables b and c, and solve for them. Then you'll know the function.
 
Dr.Peterson said:
I'm baffled. It says a=1; why are you setting it to 2? I'm also curious why they call the base of the logarithm "base" (or "bse"?) rather than using a proper variable?

As I understand the problem, they are defining a function [imath]f(x) = \log_c(bx)[/imath], and telling you that [imath]f(2) = 0[/imath] and [imath]f(3) = 1[/imath]. Plug those two pairs in to obtain two equations in variables b and c, and solve for them. Then you'll know the function.
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Could you enter the correct function?

Also, we are using bse in my classroom.

Thank you
 
Elliawt said:
Could you enter the correct function?

Also, we are using bse in my classroom.

Thank you
Click to expand...
Please make an attempt, so we can see what you have learned and where you need help. I have told you how to start.
 
Don't use the same variable name, b, for both the base and the coefficient. This is why I used c; if you want to use the name "base" for this variable, do so, though it is quite odd.

In what you did here, what happened to x? And why didn't you replace x and y with 2 and 0, then 3 and 1, as I suggested?
 
Dr.Peterson said:
Don't use the same variable name, b, for both the base and the coefficient. This is why I used c; if you want to use the name "base" for this variable, do so, though it is quite odd.

In what you did here, what happened to x? And why didn't you replace x and y with 2 and 0, then 3 and 1, as I suggested?
Click to expand...
 
I will literally send you 20$ if you solve the problem for me. I need to see your approach and final answer to understand. I have an exam on this this afternoon.. :censored:
 
It appears that you simply are not ready for the exam, and giving you an answer certainly wouldn't make you ready. You can't learn by watching others; you have to make a serious attempt, using what you have learned.

Please do exactly as I suggested. FIRST use the point (2, 0), which will immediately give you one of the parameters. THEN you can use (3, 1), which will tell you the other. What you did relates b and c; what you didn't do is needed first.
 
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