# p2 logsWatch

This discussion is closed.
#1
a very simple question...

make t the subject:

2log(3)t = m
used the change of base rule but i can't seem to think what to do with the two
0
13 years ago
#2
lnt/ln3 = m/2

(ln3)m/2 = lnt

e^((m/2)ln3) = t

e^((ln3)^(m/2)) = t

therefore, 3^(m/2) = t

Think that's right anyway....
0
13 years ago
#3
(Original post by Create)
lnt/ln3 = m/2

(ln3)m/2 = lnt

e^((m/2)ln3) = t

e^((ln3)^(m/2)) = t

therefore, 3^(m/2) = t

Think that's right anyway....
Gone round the houses there a bit!

2ln(3)t = m

ln(3)[t^2] = m

So, 3^m = t^2 and t = 3^(m/2), as you said.

Ben
0
13 years ago
#4
(Original post by Ben.S.)
Gone round the houses there a bit!
Hmm, yes, unfortunately I'm very well known for doing that! Cheers for the succinct version!
0
13 years ago
#5
breaking of with a tangent whats wrong with this method? (i have a tendancy to bend rules)

2log(3)t = m
log(3)t^2=m
3^m = t^2
mln3 = 2lnt
mln3/2 = lnt
3/2m = t
0
13 years ago
#6
(Original post by nicholls2k)
breaking of with a tangent whats wrong with this method? (i have a tendancy to bend rules)

2log(3)t = m
log(3)t^2=m
3^m = t^2
mln3 = 2lnt
mln3/2 = lnt
3/2m = t
The answer, that's what. Basically, you've just taken logs again then manipulated and 'taken them away' incorrectly. I don't see where ln3/2 comes from, unless you're meaning (1/2)ln3.

Ben
0
13 years ago
#7
How do you solve this algebraically:

y - ln x = 8
3x + y = 11
0
13 years ago
#8
yeah. The answer is y = 8 and x = 1
I am lost in my working but somehow I have arrived at (e^y) - y = (e^8) - 9

I musth ave made a mistake somewhere along the working to get 9 at the end, but if i 'fudge it' a little bit you get the right answer. Also the result is only 1 away from the other and it is a figure like 2971 = 2972. (close enough lol)
What I did is i got the two equationg and substituted the second into the first.

y - (ln(11/3 - y/3) = 8
e^y - e^ln(11/3 - y/3) = e^8
e^y - (e^ln 11/3 . e^ln y/3) = e^8
e^y - (11/3 . y/3) = e^8
e^y - 11y/9 = e^8
e^y = e^8 + 11y/9
y = ln e^8 + ln 11y/9
y = 8 + ln y - ln 9
y - ln y = 8 - ln 9
(e^y) - y = (e^8) - 9

There I get lost
0
13 years ago
#9
y-ln x=8 (i)
3x+y=11 (ii)

(i)=>y=8+lnx

(ii)=>3x+8+lnx=11=>3x+lnx=3

Raise each side to e:

x(e^(3x))=(e^(3))

Although there is no analytical way of solving this, we can equate the coefficients. On the left the coefficient of the exponential term is x while on the right it is 1. Similarly by what the 3 is multiplied on the left is x while on the right it is one, so x=1.

(i)=>y-ln1=8=>y-0=8=>y=8

=>x=1 AND y=8

Newton.
0
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