logarithm questions..

OP

Speleo

What base are your logs? I'm assuming 10.
1.) log2y = log2x + 4
log2y = log2x + log10000
log2y = log20000x
2y = 20000x
y = 10000x

That is the general method, make constants into logs, then combine the logs on each side and remove them.

sorry, base logs are not 10, subscripts didn't come up :s-smilie:

.. thanks for your help..

when i use that method i get 3/79958 for x, but the answer should be 3/22.. what am i doing wrong?

Reply 5

I think you're right, n0b0dy

Reply 6

OP

nota bene

For Q2 I guess you can use log(a/b) = log(a) - log(b)

edit: on a second thought...might not be useful...

i have tried that, and i get the wrong answer, i got stuck on log3y - 3y = 12 (log bases = 10).. how i got that? i have no idea.. :s-smilie:

Reply 7

OP

Worzo

I think you're right, n0b0dy

hmm? i doubt the answering booklet is wrong.. :s-smilie:

Reply 8

For Q2, use Alog(x) = log(x^A)

Reply 9

OP

Juwel

Or maybe the numbers between the "log" and the variable denote their bases?

what do you mean? :confused:

Reply 10

n0b0dy

hmm? i doubt the answering booklet is wrong.. :s-smilie:

Well either that or you have copied the question wrong because I get the same answer as you:

log2y = log2x + 4
log(2y/2x) = 4
y/x = 10⁴
y = 10000x

8y = 42x + 3
8*10000x = 42x + 3
(80000 - 42)x = 3
x = 3/79958
y = 30000/79958 = 15000/39979

Reply 11

Speleo

7

Are you sure they're base 10?

Reply 12

OP

Speleo

Are you sure they're base 10?

OMG OMG!! the superscripts and subscripts didn't come up!! sorry about that.. :frown:

ill put the superscripts in [].. and the subscripts in {}..

1) Solve, giving your answers as exact fractions, the simultaneous equations:
8[y] = 4[2x + 3]
log{2}y = log{2}x + 4

2)Calculate the value of y for which 2log{3}y – log{3}(y + 4) = 2

3)Calculate the values of z for which log{3}z = 4 log{z} 3

sorry again.. :s-smilie:

Reply 13

log₂(y/x) = 4 ---> y = 2⁴x ---> y = 16x

ylog₂8 = (2x+3)log₂4
3y = 4x + 6

48x = 4x + 6
44x = 6
x = 3/22

Reply 14

foxygreg

log{2}y=log{2}x+log{2}16
y=16x
8[16x]=4[2x+3]
2[3(16x)]=2[2(2x+3)]
log2[3(16x)]=log2[2(2x+3)]
48xlog2=(4x+6)log2
48x=4x+6
44x=6
x=3/22 is that right

Reply 15

OP

Worzo

log₂(y/x) = 4 ---> y = 2⁴x ---> y = 16x

ylog₂8 = (2x+3)log₂4
3y = 4x + 6

48x = 4x + 6
44x = 6
x = 3/22

seeing the 3/22 at last, thank u so much.. :smile:

Reply 16

foxygreg

3)log{3}z=log{z}81
log{z}81=log{3}81/log{3}z=81/z
81/z=z
z[2]=81
z=9

2)2log{3}y-log{3}(y+4)=2
y[2]/y+4=2
y^2-2y-8=0
(y-4)(y+2)=0
y=4, y=-2

Reply 17

OP

foxygreg

log{2}y=log{2}x+log{2}16
y=16x
8[16x]=4[2x+3]
2[3(16x)]=2[2(2x+3)]
log2[3(16x)]=log2[2(2x+3)]
48xlog2=(4x+6)log2
48x=4x+6
44x=6
x=3/22 is that right

yes, it is..

i didnt know you could just put in or take out log{2} if its on both sides.. :rolleyes:

thanks a lot for all of ur help..

Reply 18

foxygreg

no problem i haven't done logs yet but i just guess you can take them out if there on both sides i may be wrong though

Reply 19

OP