/sci/ - Science & Math

Why not, there's a log in the expression, but apart from that ... that'll be your exact value.
A decimal representation of a number is always just an approximation.

>>1054803
take some ln's dawg

>>1054812
<span class="math">0.999... \approx 1[/spoiler]

>>1054826
Oh boy, here we go...

What about : (1/(e^2))-1 for zero.

Also, I did mean the decimal approximation. How do I do it? Of course, i know how to re-arrange for exact value, but I want some nice numbers in there. thanks.

how about:

e^(1+b)=6 or e^(b)=2 ??

they're so simple but why can't i fucking do i without a calculator!

bump

Obvious homework, but who cares. Too many idiots who answer anyway.

>>1054826
Careful there, that's not the sign you're looking for.

>>1054936
I don't see why you think you should be able to, easily. Before calculators were invented they used slide rules for logarithms. Hell, I have a book that contains nothing but tables of precalculated logs of various numbers, so people wouldn't have to use slide rules.

x = 1 - (ln(5)-ln(4)) / ln(3)

ln(5/4)=ln3^(1-x)
ln(5/4)=(1-x)ln3
ln(5/4)/ln(3)=1-x
(ln(5/4)/ln(3))+1=-x
-(ln(5/4)/ln(3))+1=x
5/12+e^1=e^x

I may have gone wrong but that's sort of how to go about it i guess

<span class="math">\frac{5}{4}=3^{1-x}[/spoiler]
<span class="math">\frac{5}{4}=\frac{3}{3^x}[/spoiler]
<span class="math">5*3^x=12{/math]
<div class="math">\log_{3}^{\frac{12}{5}} = x</div>
you can simplify that further if you really want to[/spoiler]

<div class="math">\frac{5}{4}=3^{1-x}</div>
<div class="math">\frac{5}{4}=\frac{3}{3^x}</div>
<div class="math">5*3^x=12</div>
<div class="math">\log_{3}^{\frac{12}{5}} = x</div>
you can simplify further if you really want to

>>1055031
pray tell how this can be further simplified

not h/w. If i wanted solutions i would look at the worked solutions book next to me, which tells me to use my calculator.

how about: -e^(x+In2) ?

>>1055103
log 12/5 = log12 - log 5
then you can convert into natural log

>>1055119
-e^(x+ln2)=-e^(x)e^(ln2)=-2e^(x)

You can't solve for x, though, as it isn't a full expression.

>>1055125
well ok, it doesn't seem more simple to me, just different.

We would still have to use some method to approximate the logs

>>1055139
just a more standardized expression

>>1055148
mmkay

how about e^(x^-2) = x?

log3(5/4)=1-x
x= 1 - log3(5/4)

I suck at logs though so this may be erroneous

>>1055200
also forgot to put that 5/4 < 3 so its a number between 0 and 1

Somebody put this stuff in jsMath form

log (5/4) = log (3^(1-x))
log (5/4) = log (3^1 / 3^x)
log (5/4) = log 3 / log (2^x)
log(2^x)*log(5/4) = log 3
xlog2 +(5/4) = log 3
xlog2 = log 3 - (5/4)
x = (log3 -1.25 )/log2
x = log3/log2 - 5/4log2
x = log 1 - (5/log16)