 # Inverse log

I can see arc for sine, tan, etc.
But I can’t see any help for other inverse functions.

I want to find the inverse log of -6: the answer should be ~0.5

What base to the logarithmic fiunction is thought to be implied with your notation “log”?
Assuming base 10 the “inverse logarithm” of -6 should be 0.000001 .
Assuming the natural logarithm we get the result of applying the inverse function to -6 as 0.00248 (rounded!).
Assuming you meant the dyadic logarithm (base 2) the result was 0.016 (rounded).

In case of a cyclometric function the name should be derived from the name of the corresponding trigonometric function by prefixing an “arc” in the tradition of mathematics. Regrettably programming tradition has reduced the prefix to “A” (as also used with hyperbolic functions, then abbreviating “area” instead of “arcus”).

In the case of lg or ln (decadic and natural logarithm respectively) there is not such a tradition or convention.
In Calc the functions are LOG10() and LN() respectively The LOG() function allows for a second parameter to give the base and defaulting to 10 if omitted.

The inverse function with respect to ln ( LN() in Calc) is the natural exponential function exp ( EXP() in Calc).
Corresponding to LOG(Num, Base) there is not implemented an inverse under a specific name derived from “LOG”. If applied to a numeric variable Exponent you simply can use for the inverse one of the equivalent expressions `Base ^ Exponent` or `POWER(Base; Exponent)`.

There also is the equivalence `POWER(Base; Exponent) = EXP(Exponent * LN(Base) )`

unfortunately I only see exp.

That is the exponent for the base e (e=2.7182818…ish)

I need (answer) = 10^X where X is the dB/20 for an attenuator.
i.e. for a -6dB attenuator, the exponent will be -6/20 = -0.3
10^(-0.3) = ?
My calculator tells me 0.5012, which is correct

What do you miss in my answer specifically? I still think it ws rather complete.
There is not a specific function name for an exponential function to the base (10 ^(1/20)).
Simply use `=10^(-6/20)` or (somehow inefficient) `=POWER(10^(1/20); -6)`
If you have to get results for many exponents (your attenuator values) yo surely will make available the `10^(1/20)` in advance in a dedicated cell…
I do not feel well about the usage of the divisor 20 and the negative sign.

I would expect a ratio of about 0.5 for an attenuation measure of 3dB, or a gain of -3dB.

Looks like you are telling me that I should use Power() function and not exp.
I’ll try that.

6db is double or half voltage

3db is double or half power.

I see Power(,) function
it says a^b. That should do it.

Power(10,x) works, for 6dB it returns 1.995
Thanks.

OT
“6db power ratio (approximately 4 times the power / 1 fourth of the power) corresponding to a voltage ratio of double or half voltage depending on whether used for Gain / Attenuation, Common standard is to put attenuators on a dB scale by power ratio. I wanted to get the voltage ratio back without using a logarithmic scale.”?
Did I get it right?