log help (1 Viewer)

[xXmuffin0manXx]

solve for x:

i) log base 5 (1/x) = 1/2

ii )log base 2 64 = x +2

iii)log base a (x+5) - log base a (x-3) = 2


for ii)
i get 2^(x+2) = 64
x= 4

but that was from just trial and error..
is there another way ?

 

[ajdlinux]

This won't be helpful to you, but my 2U and 3U classes have done nothing on logarithms at all this year... is it even in the prelim course?

 

[shaon0]

1) 5^1/2=1/x
5^1/2=x^-1
1/2.ln(5)=-1ln(x)
-1/2.ln(5)=ln(x)
e^(-1/2.ln(5))=e^ln(x)
x=2.24

2) 2^(x+2)=64
ln(2^(x+2))=ln(64)
(x+2).ln(2)=ln(64)
x+2=ln(64)/ln(2)
x+2=6
x=4

3) a^(x+5)/(x+3) = 2
x+5 = 2x-6
x = 11

 

[tacogym27101990]

i) ln(1/x)/ln5 = 0.5
ln(1/x) = ln(rt5)
1/x= rt5
x= 1/rt5

ii) change the base again
ln64/ln2 =x+ 2

6ln(2)/ln2 =2 +x
x=4

 

[jellybelly59]

ii) (log x b)/ (log x a) = log64/log2 = x+2
= 6 = x+2
x= 4

sorry if my working out looks fuggly >.<"

 

[lyounamu]

solve for x:

i) log base 5 (1/x) = 1/2

ii )log base 2 64 = x +2

iii)log base a (x+5) - log base a (x-3) = 2


for ii)
i get 2^(x+2) = 64
x= 4

but that was from just trial and error..
is there another way ?

i) log base 5 (1/x) = (ln 1/x)/(ln 5)

i.e. (ln 1/x)/(ln 5) = 1/2
ln 1/x = ln5/2
just work it out from here.

ii) log base 2 64 = ln 64/ln 2

i.e. ln 64/ln 2 = x+2
6 = x+2
x=4

iii) log base a (x+5) - log base a (x-3) = log base a (x+5)/(x-3)
i.e. log base a (x+5)/(x-3) = 2
x+5 = 2x - 6
x = 11

 

[shaon0]

This won't be helpful to you, but my 2U and 3U classes have done nothing on logarithms at all this year... is it even in the prelim course?

yeah derivatives of logarithms are, i think.

 

[xXmuffin0manXx]

ohh..i see

just a few little concepts that i had muddled up..

thanks for all your help !

edit: looks like i got marked wrong for the third one too

 

[lolokay]

solve for x:

i) log base 5 (1/x) = 1/2

ii )log base 2 64 = x +2

iii)log base a (x+5) - log base a (x-3) = 2


for ii)
i get 2^(x+2) = 64
x= 4

but that was from just trial and error..
is there another way ?

for ii) you should know your powers of 2, but to solve it without trial and error just put ln 64/ ln 2 into the calculator, which gives you 6

for i) another way to do it would be to use the fact that
log₅[1/x] = -log₅x
which gives x = 5^-1/2