Logs
Given that log2X =a, find in terms of a, the simplest form of:
a) Log2(16x)
b) log2(x^4/2)
c) Hence or otherwise solve
Log2(16x)-log2(x^4/2) = 1/2
giving your answer in its simplest surd form.
a) Log2(16x)
b) log2(x^4/2)
c) Hence or otherwise solve
Log2(16x)-log2(x^4/2) = 1/2
giving your answer in its simplest surd form.
a) log2(16x) = log2(8) + log2x = lg8/log2 + a = 3 + a
b) log2(x^4/2) = log2(x^4) - log2(2) = 4log2x - lg2/lg2 = 4a - 1
=> 3+a - (4a-1) = 1/2
=> 3 + a - 4a + 1 = 1/2
=> 4 - 3a = 1/2
=> 3a = 7/2
=> a = 7/6
log2x = 7/6
lgx/lg2 = 7/6
lgx = 7/6 * lg2
x = 10^(7/6 * lg2)
I probably messed it up somewhere, but it should help you somewhat.
b) log2(x^4/2) = log2(x^4) - log2(2) = 4log2x - lg2/lg2 = 4a - 1
=> 3+a - (4a-1) = 1/2
=> 3 + a - 4a + 1 = 1/2
=> 4 - 3a = 1/2
=> 3a = 7/2
=> a = 7/6
log2x = 7/6
lgx/lg2 = 7/6
lgx = 7/6 * lg2
x = 10^(7/6 * lg2)
I probably messed it up somewhere, but it should help you somewhat.
Hello!
Um... I ended up with
Log(16x) = log16 + logx = log(2^4) + a = 4 log2 + a = 4 + a
(because log 2 is 1 if you are using logs in base 2 – sorry, can’t get the editor to take subscripts)
And
Log (x^4/2) = log (x^4) – log 2 = 4 log x - log 2 = 4 log x – 1 = 4a – 1
Combining these gives
Log(16x) – log (x^4/2) = 4 + a – (4a – 1) = 4 + a – 4a + 1 = 5 – 3a
Now the question says that this is 1/2, so
5 – 3a = 1/2, so 3a = 9/2 (aka 4.5) and so a = 3/2
So if log x = a, and a = 3/2,
then log x = 3/2 and so x = 2^(3/2)
so x = the square root of 8 aka 2root2 (in its simplest surd form)
That any use?
love danni
PS Knogle, sorry but when you split the log(16x) into log 8 + log (2x) you'd either misread the 16x as 8x or decided the 2 in 2x was the base not the, er, thing you were taking the log of (if that makes sense)
Um... I ended up with
Log(16x) = log16 + logx = log(2^4) + a = 4 log2 + a = 4 + a
(because log 2 is 1 if you are using logs in base 2 – sorry, can’t get the editor to take subscripts)
And
Log (x^4/2) = log (x^4) – log 2 = 4 log x - log 2 = 4 log x – 1 = 4a – 1
Combining these gives
Log(16x) – log (x^4/2) = 4 + a – (4a – 1) = 4 + a – 4a + 1 = 5 – 3a
Now the question says that this is 1/2, so
5 – 3a = 1/2, so 3a = 9/2 (aka 4.5) and so a = 3/2
So if log x = a, and a = 3/2,
then log x = 3/2 and so x = 2^(3/2)
so x = the square root of 8 aka 2root2 (in its simplest surd form)
That any use?
love danni
PS Knogle, sorry but when you split the log(16x) into log 8 + log (2x) you'd either misread the 16x as 8x or decided the 2 in 2x was the base not the, er, thing you were taking the log of (if that makes sense)
I knew i should have worked it out on paper. 
)
=> Combining these gives
Log(16x) – log (x^4/2) = 4 + a – (4a – 1) = 4 + a – 4a + 1 = 5 – 3a
Now the question says that this is 1/2, so
*=> 5 – 3a = 1/2, so 3a = 9/2 (aka 4.5) and so a = 3/2
**=> 5 - 3a = 1/2, so 3a = -5 + 1/2, 3a = -9/2 and so a = a = -9/6 ~ -3/2
So if log2 x = a, and a = -3/2,
then log2 x = -3/2 and so x = 2^(-3/2) => x = 1 / 2^(3/2)
so x = 1 / square root of 8 aka 2 root 2 (in its simplest surd form)
Log(16x) – log (x^4/2) = 4 + a – (4a – 1) = 4 + a – 4a + 1 = 5 – 3a
Now the question says that this is 1/2, so
*=> 5 – 3a = 1/2, so 3a = 9/2 (aka 4.5) and so a = 3/2
**=> 5 - 3a = 1/2, so 3a = -5 + 1/2, 3a = -9/2 and so a = a = -9/6 ~ -3/2
So if log2 x = a, and a = -3/2,
then log2 x = -3/2 and so x = 2^(-3/2) => x = 1 / 2^(3/2)
so x = 1 / square root of 8 aka 2 root 2 (in its simplest surd form)
Original post by annikapaula
...
7 years old and you bump this thread
Poor form
Original post by danniella
Hello!
Um... I ended up with
Log(16x) = log16 + logx = log(2^4) + a = 4 log2 + a = 4 + a
(because log 2 is 1 if you are using logs in base 2 – sorry, can’t get the editor to take subscripts)
And
Log (x^4/2) = log (x^4) – log 2 = 4 log x - log 2 = 4 log x – 1 = 4a – 1
Combining these gives
Log(16x) – log (x^4/2) = 4 + a – (4a – 1) = 4 + a – 4a + 1 = 5 – 3a
Now the question says that this is 1/2, so
5 – 3a = 1/2, so 3a = 9/2 (aka 4.5) and so a = 3/2
So if log x = a, and a = 3/2,
then log x = 3/2 and so x = 2^(3/2)
so x = the square root of 8 aka 2root2 (in its simplest surd form)
That any use?
love danni
PS Knogle, sorry but when you split the log(16x) into log 8 + log (2x) you'd either misread the 16x as 8x or decided the 2 in 2x was the base not the, er, thing you were taking the log of (if that makes sense)
Um... I ended up with
Log(16x) = log16 + logx = log(2^4) + a = 4 log2 + a = 4 + a
(because log 2 is 1 if you are using logs in base 2 – sorry, can’t get the editor to take subscripts)
And
Log (x^4/2) = log (x^4) – log 2 = 4 log x - log 2 = 4 log x – 1 = 4a – 1
Combining these gives
Log(16x) – log (x^4/2) = 4 + a – (4a – 1) = 4 + a – 4a + 1 = 5 – 3a
Now the question says that this is 1/2, so
5 – 3a = 1/2, so 3a = 9/2 (aka 4.5) and so a = 3/2
So if log x = a, and a = 3/2,
then log x = 3/2 and so x = 2^(3/2)
so x = the square root of 8 aka 2root2 (in its simplest surd form)
That any use?
love danni
PS Knogle, sorry but when you split the log(16x) into log 8 + log (2x) you'd either misread the 16x as 8x or decided the 2 in 2x was the base not the, er, thing you were taking the log of (if that makes sense)
Hey, uhm Thanks a lot for the detailed answer 17 years later who would of figured this would help someone, funny thing is the question is older than me!! I'm only 17 this year haha. still really appreciate it Danni hope you see this. Peace Moodi
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