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Posted

Heyo,

 

Anyone aware of a calculator out there on the internet with capacity to calculate logarithms with non-common bases?

 

i.e.

 

log(base2)4(16^(1/3)) - log(base4)0.25^(1/3)

 

= x

 

 

Yes, yes, I know x=11/3

 

I just want a quick way to check my answers.

 

Alternatively, is there a quick and easy way to convert non-common bases to common bases? (base10)

Posted (edited)

Anyone aware of a calculator out there on the internet...

No matter what the question here is, the answer is Wolfram Alpha.
Alternatively, is there a...
Aaand when you just need to look something up, Wikipedia. Edited by the tree
Posted

Alternatively, is there a quick and easy way to convert non-common bases to common bases? (base10)

Wikipedia sais [math] \log_b r = \frac{\log_a r}{\log_a b} [/math]

Posted

Okay, erm, having some trouble.

 

Here's the question:

 

An investment pays 6%/annum

Principle=$3000

Final=$6000

Time= (?)

 

Here's what I keep getting, but something seems very, very wrong:

 

F = P (1+i)^t

 

$6000 = $3000 (1.06)^t

$2000 = 1.06^t

 

one way...

t = log_1.06(2000)

t = (log2000)/(log1.06)

t = 130.445

 

another way...

log2000 = log1.06^t

log2000 = (t) log1.06

log2000/log1.06 = 130.445

 

What am I doing wrong here? There's no way it takes 130 years, 5 months, ~10 days, 4 hours, and 48 minutes to double an investment at 6%/annum.

Posted

Bah, I seem to have developed a mental block around logs... can someone tell me if this is right?

 

 

M=magnitude

I=Intensity

U=Baseline

 

(M = log (I/U) )

 

An earthquake registers 6.1 on the Richter Scale.

What is the rating on a quake twice as powerful?

 

 

 

My answer is :

 

Let "M1" be the quake with a magnitude of 6.1.

Let "M2" be the unknown quake magnitude.

Let the intensity of M2 be called "I1"

Let the intensity of the unknown quake be called "I2"

(M1 = log (I/U) )

M1 = log (I/U)

I/U = 10^(M1)

I = ( 10^(M1) ) * (U)

(I2/I1) = 2

 

The ratio of these two values must = 2 (for a quake twice as intense).

 

(I2/I1) = (( 10^M2 )*( U )) / (( 10^M1 )*( U ))

 

The baseline values ("U") divide out.

 

(I2/I1) = (10^M2 ) / ( 10^M1 )

 

Substitute value for "M1"

 

(I2/I1) = (10^M2 ) / ( 10^(6.1) )

(I2/I1) = 10^(M2 - 6.1)

 

Substitute value for (I2/I1)

 

2 = 10^(M2 - 6.1)

 

AKA (in common log form):

 

M2 - 6.1 = log2

 

Calculator...

 

( M2 - 6.1 ) = 0.301029996

M2 = ( 0.301029996 + 6.1 )

M2 = 6.401029996

 

~ to sig. digits

 

M2 = 6.4

 

 

In retrospect I should've just found:

 

10^(x)=2

where:

x=(M2-M1)

 

 

 

 

I seem to be having trouble with logs. Does anyone have some tips to avoid over complicating these problems?

 

 

Posted

I don't really understand the question, because I could only guess which variables correspond to "rating" and "power". Anyways, perhaps the identity "log(2x) = log 2 + log x" helps you? Another related and helpful one is [math] \log a^n = n \, \log a [/math], btw. Should both work in all bases.

Posted

well, M is the magnitude, the richter value

the question was non-specific... so yeah. i assume its referring to intensity because the richter scale is an exponential function

  • 2 weeks later...
Posted

Hi sysD.

Yes Wolfram Alfa is amazing.

I wrote a little on-line calculator myself. Nothing fancy but does what I need better than any other I could find.

It happens to be able to solve your logarithm problem: pascalc.com -you should see the solution by clicking on the link.

Posted (edited)

In the calculator "log" calculation is carried out by using power series.

The problem occurs very small long digit calculation.

In this case if you need more accurate value, you had better use more modified method.

Edited by alpha2cen
Posted

Hi sysD.

Yes Wolfram Alfa is amazing.

I wrote a little on-line calculator myself. Nothing fancy but does what I need better than any other I could find.

It happens to be able to solve your logarithm problem: pascalc.com -you should see the solution by clicking on the link.

 

Haha, thanks man. This is excellent. *bookmarked*

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