Board 8 > 3^x = 8?

(+) Yeah! (+)

How do you solve for x here without a calculator? This was on a test I had today..

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Take the natural log (or regular log, I guess) of both sides.

x * ln(3) = ln(8)

x = ln(8)/ln(3)

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without a calculator? You don't. Well, not unless you've got a slide rule handy...

Well, yeah, if you want a decimal value of that... yeah, that's not happening in most cases.

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TheFlyingWolfen posted...
How do you solve for x here without a calculator? This was on a test I had today..

Are you sure the question wasn't 2^x = 8?

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was it multiple choice? if so what were the possibilities?

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I worked it out (with a calculator) and the answer is X= 1.892789261

Good luck finding that without a calculator.

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Chrono1219

just memorize log tables.

log(2) = .301
log(3) = .477

take the logarithm of your original problem, you get:
x * log(3) = log(8)
x * log(3) = 3 * log (2)
x = 3 * log(2) / log(3)
x = .903/.477
x ~ 1.9 (too lazy to do more accurately using mental math)

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guesstimate imo

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Chrono1219 posted...
I worked it out (with a calculator) and the answer is X= 1.892789261

Good luck finding that without a calculator.

Well, it's obviously -close- to that.

3^2 = 9.
3^1 = 3.

Even if you assume a linear interpolation, 8 is 5/6 of the way from 3 to 9, so you'd get...

X ~= 1.83

As a first guess based on a totally ridiculous assumption of linearity.

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In all seriousness, though, I suppose you could use the Taylor expansion for ln(x)....

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Cats land on their feet. Toast lands peanut butter side down. A cat with toast strapped to its back will hover above the ground in a state of quantum indecision

You can't really find a decimal value without a calculator or memorizing tables. But I'm guessing either that's not what the question actually was, or you had some instructions other than actually solving for the exact value of x, or your teacher would accept something like log(8)/log(3) or a series expansion or something specific to the class. Especially if it's a higher-level course where the actual computation is pretty much totally meaningless since you just use a calculator/computer for that part and focus on concepts and thinking skills.

Edit: Or it could be multiple-choice and you're supposed to recognize that it's close to but not quite 2 and pick the appropriate answer.

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Fun stuff here:

http://www.curiousmath.com/index.php?name=News&file=article&sid=32&theme=Printer

This is similar to the above ("just memorize the log table") but there's some good points.

In particular, the point about only memorizing prime numbers, and then getting the composite numbers by adding prime numbers together. If you want one digit of accuracy, you only really need to memorize three logarithms.

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Cats land on their feet. Toast lands peanut butter side down. A cat with toast strapped to its back will hover above the ground in a state of quantum indecision

Really, you only have to memorize three or four key values:

log(2) = .301 (easy to remember: 2^10 is 1024 is roughly 10^3, so log(2)~3/10)
log(6) = .778 (7/9, so easy to remember)
log(7) = .845 (tough to remember, but less important than the first two)
log(e) = .434 (again, tough but less important).

From there, log(5) = log(10/2) = log(10) - log(2) = 1 - .301 = .699
log(3) = log(6/2) = log(6) - log(2) = .778 - .301 = .477
For everything else, either combine or just use linear interpolation of nearby points (it starts working fairly accurately once you get above 10--just remember to round up on everything).

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the_rowan posted...
You can't really find a decimal value without a calculator or memorizing tables.

You can; I mean logs have been around since the 1600s. Obviously someone had to make the log tables.

Here, let me start grinding some stuff out by hand.

2^10 = 1024, which is really close to 10^3

therefore
10 * log 2 ~= 3

therefore
log 2 ~= 3/10 = 0.30
(actual value: 0.3010...)

3^2 = 9, which is close to 10.

Therefore
2 * log 3 ~= 1
log 3 ~= 0.5
(actual value: 0.4771...)

Or, and admitedly, I didn't calculate this by hand:
3^21 = 10,460,353,203

Therefore
21 * log 3 ~= 10
log 3 ~= 10/21 = 4.76 (and that I did calculate by hand)

Log 4 = 2 * log 2 ~= 0.60
(actual value: 0.6020)

log 5 = log 10 - log 2 ~= 0.70
(actual value: .6990)

log 6...well I could just add the log of 2 and 3 together, but I cheated a little on log 3, so let's see...

6^1 = 6
6^2 = 36
6^3 = 216
6^4 = 1296
6^5 = 1300*6 - 4*6 = 7800 - 24 = 7776
6^6 = 7777*6 - 6 = 46662 - 6 = 46656

Ok, I don't want to compute much further by hand; the one that looks the most useful at a glance is 6^3 = 216

3 * log 6 ~= log(200) = 2 + log 2 ~= 2.3

log 6 ~= 2.3/3 = 2/3 + 0.1 = 0.76666...
(actual value: 0.7781)

Log 7...

Right away, 7^2 = 49 ~= 50 jumps to mind.

2 * log 7 ~= log(50) = 1 + log 5 ~= 1.70
log 7 ~= 0.85
(actual value: .8451)

Log 8...

Log 8 = 2 * log 2 ~= 0.90
(actual value 0.9031)

Log 9...

Well...I did cheat a bit on the log 3 stuff, so let me go back and use the log 6 result to re-calculate log 3.

log 3 = log 6 - log 2 ~= 0.77 - 0.30 = 0.47
(Actual value: 0.4771)

Which means...

Log 9 = 2 * Log 3 ~= 0.94
(Actual value: .9542)

So there you go: all the one-digit logarithms hand calculated to *almost* 2 digits of accuracy (I was off by 0.1 in a couple of places--3, 6, and 9. Damn you multiples of 3 *shakes fist*).

Actually, let's see if I can get a better estimate on log 3/6/9. I know 12^2 is 144. I know the square root of 2 is 1.41.... So...without even calculating it, I know that 12^4 is going to be pretty close to 20000 (closer than 216 is to 200, anyhow).

4 * log 12 ~= 4 + log 2

log 12 ~= 1 + 0.30/4 ~= 1.08
(actual value 1.0791)

Which gives us...
Log 3 = Log 12 - Log 4 ~= 1.08 - 0.60 = 0.48
(actual value: 0.4771)

Log 6 = Log 12 - Log 2 ~= 1.08 - 0.30 = 0.78
(actual value: 0.7781)

Log 9 = 2 * Log 3 ~= 0.96
(actual value: 0.9542)

Arg, nooo, I'm still off.

Wait, 81, is pretty close to 80.

2 * log 9 ~= 1 + log 8 ~= 1.90
log 9 ~= 0.95
(actual value: 0.9542)

There--now I've gotten estimates for every one-digit logarithm that's accurate up to two digits. All calculated by hand. Yaaay.

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From: metroid composite | #013
In all seriousness, though, I suppose you could use the Taylor expansion for ln(x)....

I was gonna say this but it doesn't work for x > 1

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You guys just couldn't leave it at ln(8)/ln(3) could you?

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From: tyder21 | #019
You guys just couldn't leave it at ln(8)/ln(3) could you?

Hey, I learned something!

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If you guys wanted to get all crazy you should've used Newton's method for solving 3^x-8 = 0.

Derivative of 3^x - 8 = 3^xln(3)

A decent guess for x is 2.

2 - (3^2-8)/(3^2ln(3)) = 2 - 1/(9ln(3))
Since e is almost 3, 1/ln(3) is almost 1.
2 - (1/9)
17/9 = 1.8~

Memorizing the log table is lame.

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So, an inquiring reader might ask, how would I go further? How would I get, say, three digits of accuracy? And how, without looking up on the internet, would I know that I had three digits of accuracy?

Ok, let's cover accuracy first.

You'll notice that when we were very close to powers of ten (2^10 = 1024) that we were quite accurate. Notably, 1024 is only 2.4% away from being 10^3, and 0.30 is actually even more accurate--0.34% away from being log(2). We can predict this with a good degree of accuracy using Calculus.

The derivative of Log( x ) at any one point is...well Log( x ) = ln(x) / ln(10), so the derivative is 1/x * 1/ln(10). I don't even need to know ln(10) here, just that it's somewhere between 2 and 3 (cal it 2.5).

So, this means that at x = 1000, the slope is roughly 1/1000 * 1/(2.5) = 1/2500. Going 24 away from this, we'd expect the answer to be off by 24/2500, or ~0.01. And indeed, log(1000) = 3. log(1024) = 3.0103. (Actually...we'd actually be freakishly accurate here if we had a better estimate on ln(10) than 2.5. My initial plan had just been to show that we could get a pretty good guess on the error, but this might be worth pursuing...).

Man, I was going to go calculate the square root of ten, cube root of ten, etc by hand, but now I'm tempted to screw that and just get a really good estimate on log(e), and trusting in the power of calculus. Well...square root of ten is still going to come in handy, and is honestly probably easier to calculate by hand than e.

Ok, square root of 10, let's do this!

Pass 1:
3*3 = 9
4*4 = 16
First digit is 3, second digit is small.

Pass 2:
3.1*3.1 = 9.61
3.2*3.2 = 10.24
Second digit is 3, third digit is largish.

Pass 3:
3.17*3.17 = 3.1*3.1 + 2*(3.1*0.07) + (0.07)^2 = 9.61 + 2*(0.217) + 0.0049 > 9.61 + 0.43 + 0 = 10.04
3.16*3.16 = 3.1*3.1 + 2*(3.1*0.06) + (0.06)^2 = 9.61 + 2*(0.186) + 0.0036 = 9.61 + 0.372 + 0.0036 = 9.9856
Third digit is 6, fourth digit is smallish.

Hmm...well, let me state up front that I'm not going to calculate the fifth digit because this is becoming a pain, so that lets me cut a couple of corners here
3.163*3.163 ~= 9.9856 + 2*(3.16*0.003) ~= 9.9856 + 2*(0.009) ---- (Nope--too big. That's adding 0.018. We're looking for 0.0144)
3.162*3.162 ~= blah blah + 2*(3*0.002) ~= blah blah + 0.012 ---- (Yep, that's better--closer to the 0.0144)
(Fourth digit is 2, fifth digit looks small)

So to four digits of accuracy, sqrt(10) = 3.162

Ok, how much of a pain is it going to be to compute e?

e = sum( 1/n! ) from n = 0 to infinity
e = 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 + ...
~= 2.5 + (20 + 5 + 1)/120 = 2.5 + 13/60 = 2.5 + 1.3/6 = 2.5 + 1/6 + 0.3/6 = 2.5 + 1/6 + 0.05 = 2.55 + 1/6 ~= 2.55 + 0.16667 = 2.7167

The next term is 1/720, which we can totally ballpark as roughly 1/700 ~= 0.0014. So...that brings us up to 2.7181

The next term is 1/5040, which we can totally ballpark as roughly 1/5000 ~= 0.0002. So...that brings us up to 2.7183

And the term after that is going to be too small to register at 5 digits of accuracy. Bam, 5-digits of accuracy on e. That was remarkably painless. Man, four digits on the square root of 10 hurt more than that.

Ok, just a ballpark, though, I'm going to take it back down to 2.7.

e^2 ~= 4 + 2*(2*0.7) + 0.49 = 4.49 + 2.8 = 7.29 ~= 7.3
With three digits..
e^2 ~= 7.29 + 2*(0.02*2.7) + something_tiny = 7.29 + (0.108) ~= 7.4

Well, at very least this gives us a very rough bead. We know to a good degree of accuracy, that log(8) = 0.90, and log(7) = 0.85, so let's say that log(7.4) = 0.87. That means log(e) is pretty close to 0.435, which in turn means that ln(10) is pretty close to 1/0.435.

(4k check; time to break)

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Cats land on their feet. Toast lands peanut butter side down. A cat with toast strapped to its back will hover above the ground in a state of quantum indecision