MedVision ad

logs? (1 Viewer)

darshil

Replicant
Joined
Mar 20, 2008
Messages
234
Location
my mind
Gender
Undisclosed
HSC
2009
i coudnt do it, appreciating everyone's help
(log base 10 = log unless the base is other than that)

(1). 2logx + 3 = log(x^5) -> blank on this
(2). log (1+root5 / 2) + log (3+ root5 / 2) -> rationalise?
All your help is always appreciated.
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
i coudnt do it, appreciating everyone's help
(log base 10 = log unless the base is other than that)

(1). 2logx + 3 = log(x^5) -> blank on this
(2). log (1+root5 / 2) + log (3+ root5 / 2) -> rationalise?
All your help is always appreciated.
(1). 2logx + 3 = log(x^5)
2log x + 3 = log (x^5)
log x^2 + log 10^3 = log (x^5)

log (1000x^2) = log (x^5)
x^5 = 1000 x ^2

x^5 - 1000x^2 = 0
x^2(x^3 - 1000) = 0
x = 0 or x = 10
BUT x =/= 0, so x = 10

(2). log (1+root5 / 2) + log (3+ root5 / 2) -> rationalise?

log (((1+root5)/2) x (3+root5)/2)) = log ((8+4root5)/4) = log (2+root5)
 

Xcelz

Member
Joined
Nov 6, 2008
Messages
96
Gender
Male
HSC
2009
i coudnt do it, appreciating everyone's help
(log base 10 = log unless the base is other than that)

(1). 2logx + 3 = log(x^5) -> blank on this
(2). log (1+root5 / 2) + log (3+ root5 / 2) -> rationalise?
All your help is always appreciated.
1. 2logx + 3 = log(x^5)
2logx + 3 = 5logx
3logx=3 (subtracted 2logx from each side)
logx=1 (divided each side by 3)
x=10

2. log (1+root5 / 2) + log (3+ root5 / 2)

= log((1+root5/2) x (3+root5/2)) (You have to expand the brackets and multiply the denominators)
=log((3+root5+3root5+5)/4)
=log ((8+4root5)/4) (divide everything by 4)
=log (2+root5)

Hope that helps
 
Last edited:

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
1. 2logx + 3 = log(x^5)
2logx + 3 = 5logx
3logx=3 (subtracted 2logx from each side)
logx=1 (divided each side by 3)
x=1
Nearly correct.

But it's log base 10.

So if logx = 1,

x = 10.
 

darshil

Replicant
Joined
Mar 20, 2008
Messages
234
Location
my mind
Gender
Undisclosed
HSC
2009
Thanks guys, really appreciate it,
all the best !
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top