Logarithms
Hello,
Could someone please help me with this problem step by step - not sure how tou get to the answer - can only get so far and :-(
11a) Show that log[4]3 = log[2]sqrt3
11b) Hence or otherwise solve the simultaneous equation:
2log[2]y = log[4]3 + log[2]x
3^y = 9^x
Thanks for you help!
Could someone please help me with this problem step by step - not sure how tou get to the answer - can only get so far and :-(
11a) Show that log[4]3 = log[2]sqrt3
11b) Hence or otherwise solve the simultaneous equation:
2log[2]y = log[4]3 + log[2]x
3^y = 9^x
Thanks for you help!
Original post by christinajane
Hello,
Could someone please help me with this problem step by step - not sure how tou get to the answer - can only get so far and :-(
11a) Show that log[4]3 = log[2]sqrt3
11b) Hence or otherwise solve the simultaneous equation:
2log[2]y = log[4]3 + log[2]x
3^y = 9^x
Thanks for you help!
Could someone please help me with this problem step by step - not sure how tou get to the answer - can only get so far and :-(
11a) Show that log[4]3 = log[2]sqrt3
11b) Hence or otherwise solve the simultaneous equation:
2log[2]y = log[4]3 + log[2]x
3^y = 9^x
Thanks for you help!
How did you get on with 11a?
Original post by SeanFM
How did you get on with 11a?
I got, which I don't even know if its right :
log[4]3 = 1/2log[2]3 = log[2]3^1/2 = log[2]sqrt3
Which seems too obvious????
Original post by christinajane
I got, which I don't even know if its right :
log[4]3 = 1/2log[2]3 = log[2]3^1/2 = log[2]sqrt3
Which seems too obvious????
log[4]3 = 1/2log[2]3 = log[2]3^1/2 = log[2]sqrt3
Which seems too obvious????
It's correct! Keep going.
Oops forgot first line - log[4]3 = log[2]3/log[2]4 = log[2]3/2
but dont really now why that eqals log[2]sqrt3??
but dont really now why that eqals log[2]sqrt3??
Original post by christinajane
I got, which I don't even know if its right :
log[4]3 = 1/2log[2]3 = log[2]3^1/2 = log[2]sqrt3
Which seems too obvious????
log[4]3 = 1/2log[2]3 = log[2]3^1/2 = log[2]sqrt3
Which seems too obvious????
There is a rule you can use to change the base of a logarithm.
loga(b)=[logc(b)]/[logc(a)].
(edited 8 years ago)
Second part I get -
2log[2]y = log[4]3 + log[2}x 3^y = 9^x
y = 2^x (2)
log[2]y^2 = log[2]sqrt3 + log[2]x
y^2 = x(sqrt3) (1)
It follows that:
substituting equation 2 into 1
(2^x)^2 = xsprt3
4x^2 = xsqrt3
then I get stuck...
2log[2]y = log[4]3 + log[2}x 3^y = 9^x
y = 2^x (2)
log[2]y^2 = log[2]sqrt3 + log[2]x
y^2 = x(sqrt3) (1)
It follows that:
substituting equation 2 into 1
(2^x)^2 = xsprt3
4x^2 = xsqrt3
then I get stuck...
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