MapInfo Pro

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hi Peter,

The results don't look correct? These look to be LN, not log ₁₀ (x). I checked a couple on my calculator and the results in MIP are not Log ₁₀ (x)

Results I'm getting are different to what I'm expecting Peter. My mistake if I didn't specify Log ₁₀ (x) = Log

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hi Peter,

The results don't look correct? These look to be LN, not log ₁₀ (x). I checked a couple on my calculator and the results in MIP are not Log ₁₀ (x)

Results I'm getting are different to what I'm expecting Peter. My mistake if I didn't specify Log ₁₀ (x) = Log

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hi Peter,

The results don't look correct? These look to be LN, not log ₁₀ (x). I checked a couple on my calculator and the results in MIP are not Log ₁₀ (x)

Results I'm getting are different to what I'm expecting Peter. My mistake if I didn't specify Log ₁₀ (x) = Log

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hi Peter,

The results don't look correct? These look to be LN, not log ₁₀ (x). I checked a couple on my calculator and the results in MIP are not Log ₁₀ (x)

Results I'm getting are different to what I'm expecting Peter. My mistake if I didn't specify Log ₁₀ (x) = Log

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hi Peter,

As mentioned, your suggestion for calc the log base 10 of my Au ppb works using; Au_ppb, Log(Au_ppb)/Log(10)

I'm just confused, as the equations implies the division by Log (10) [Log (10) = 1], so I was unsure how MIP is calculating the correct value = Ln/1

Cheers, Mike

Hi Peter,

The results don't look correct? These look to be LN, not log ₁₀ (x). I checked a couple on my calculator and the results in MIP are not Log ₁₀ (x)

Results I'm getting are different to what I'm expecting Peter. My mistake if I didn't specify Log ₁₀ (x) = Log

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hi Peter,

As mentioned, your suggestion for calc the log base 10 of my Au ppb works using; Au_ppb, Log(Au_ppb)/Log(10)

I'm just confused, as the equations implies the division by Log (10) [Log (10) = 1], so I was unsure how MIP is calculating the correct value = Ln/1

Cheers, Mike

Hi Peter,

The results don't look correct? These look to be LN, not log ₁₀ (x). I checked a couple on my calculator and the results in MIP are not Log ₁₀ (x)

Results I'm getting are different to what I'm expecting Peter. My mistake if I didn't specify Log ₁₀ (x) = Log

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hi Peter,

Yes, Log Base_-10. I have included some examples of Au ppb column and the MIP syntax from Friday

The calc/results in MIP are correct;

(Au_ppb)/Log(10) = 2.70805/ 2.3026 (log of 10) = 1.17609 [Log_-10 (15)

If MIP could please add in functionality for Log_-10 this would be v. helpful

Thank you for your help Peter!

Cheers

Hi Peter,

As mentioned, your suggestion for calc the log base 10 of my Au ppb works using; Au_ppb, Log(Au_ppb)/Log(10)

I'm just confused, as the equations implies the division by Log (10) [Log (10) = 1], so I was unsure how MIP is calculating the correct value = Ln/1

Cheers, Mike

Hi Peter,

The results don't look correct? These look to be LN, not log ₁₀ (x). I checked a couple on my calculator and the results in MIP are not Log ₁₀ (x)

Results I'm getting are different to what I'm expecting Peter. My mistake if I didn't specify Log ₁₀ (x) = Log

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hi Peter,

Yes, Log Base_-10. I have included some examples of Au ppb column and the MIP syntax from Friday

The calc/results in MIP are correct;

(Au_ppb)/Log(10) = 2.70805/ 2.3026 (log of 10) = 1.17609 [Log_-10 (15)

If MIP could please add in functionality for Log_-10 this would be v. helpful

Thank you for your help Peter!

Cheers

Hi Peter,

As mentioned, your suggestion for calc the log base 10 of my Au ppb works using; Au_ppb, Log(Au_ppb)/Log(10)

I'm just confused, as the equations implies the division by Log (10) [Log (10) = 1], so I was unsure how MIP is calculating the correct value = Ln/1

Cheers, Mike

Hi Peter,

The results don't look correct? These look to be LN, not log ₁₀ (x). I checked a couple on my calculator and the results in MIP are not Log ₁₀ (x)

Results I'm getting are different to what I'm expecting Peter. My mistake if I didn't specify Log ₁₀ (x) = Log

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hi Peter,

Yes please!! Please add a function that we can specify the base as a second parameter, perhaps into one of the functions in the query setup.

Appreciated.

Cheers

Hi Peter,

Yes, Log Base_-10. I have included some examples of Au ppb column and the MIP syntax from Friday

The calc/results in MIP are correct;

(Au_ppb)/Log(10) = 2.70805/ 2.3026 (log of 10) = 1.17609 [Log_-10 (15)

If MIP could please add in functionality for Log_-10 this would be v. helpful

Thank you for your help Peter!

Cheers

Hi Peter,

As mentioned, your suggestion for calc the log base 10 of my Au ppb works using; Au_ppb, Log(Au_ppb)/Log(10)

I'm just confused, as the equations implies the division by Log (10) [Log (10) = 1], so I was unsure how MIP is calculating the correct value = Ln/1

Cheers, Mike

Hi Peter,

The results don't look correct? These look to be LN, not log ₁₀ (x). I checked a couple on my calculator and the results in MIP are not Log ₁₀ (x)

Results I'm getting are different to what I'm expecting Peter. My mistake if I didn't specify Log ₁₀ (x) = Log

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hi Peter,

The results don't look correct? These look to be LN, not log ₁₀ (x). I checked a couple on my calculator and the results in MIP are not Log ₁₀ (x)

Results I'm getting are different to what I'm expecting Peter. My mistake if I didn't specify Log ₁₀ (x) = Log

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike

Hello Mike,

The MapBasic log() function is for natural (base e) logs. You can confirm this by taking the log of e itself, which should yield 1.

print log(2.718281828459)

outputs: 1

To find the log of another base, such as 10, simply divide the natural log of your number by the natural log of the desired base.

print log(2.718281828459)/log(10)

outputs: 0.434294

Hopefully these match both your calculator and your expectations!

In you're working in Python, the math module has log2(x), log10(x) and log(x, base) functions.

Regards,

Warren Vick

Hi Peter,

The results don't look correct? These look to be LN, not log ₁₀ (x). I checked a couple on my calculator and the results in MIP are not Log ₁₀ (x)

Results I'm getting are different to what I'm expecting Peter. My mistake if I didn't specify Log ₁₀ (x) = Log

Cheers, Mike

Hi Peter,

Perfect, it worked. Thank you Peter

Is there a way to write the LogAu, back to the original table, placing the Log(Au_ppb) at the end of the original table?

I take it you could also apply logs to all e.g. elements in the table, in 1 step?

Cheers, Mike