# Maths logs

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#1
Can some explain how i am supposed to complete this question?
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#2

6i)
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6 months ago
#3
you have to use laws of logs. log(an)=nloga
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6 months ago
#4
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#5

thanks i got that im stuck on this one now
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6 months ago
#6
(Original post by A0W0N)

thanks i got that im stuck on this one now
Which one?
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#7
7ii)
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#8

i don't don't how to find x
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6 months ago
#9
(Original post by A0W0N)

i don't don't how to find x
you can remove logs from both sides because you know n0 = 1, where n =/= 0
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6 months ago
#10
(Original post by A0W0N)

i don't don't how to find x
Tbh it would be easier to take the log(7) to the right hand side and note
63 = 9*7
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#11

why is it telling me to use base 10?
(i put what i think the answer is on the calculator)
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6 months ago
#12
(Original post by A0W0N)

why is it telling me to use base 10?
(i put what i think the answer is on the calculator)
The right hand side are (mostly) easy to calculate using base 10.
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#13
(Original post by mqb2766)
The right hand side are (mostly) easy to calculate using base 10.
but wouldn't they produce different results?
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6 months ago
#14
(Original post by A0W0N)
but wouldn't they produce different results?
Of course. Using base 10
log(0.01) = -2 as 10^(-2) = 0.01
log(0.1) = -1 as 10^(-1) = 0.1
log(1) = 0 as 10^0 = 1
log(10) = 1 as 10^1 = 10
log(100) = 2 as 10^2 = 100
..
Last edited by mqb2766; 6 months ago
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6 months ago
#15
(Original post by A0W0N)

why is it telling me to use base 10?
(i put what i think the answer is on the calculator)
You’ve used base 2. They want base 10 so you can only taken logarithms in base 10.
So x=log(10^6)/log(2) which gives the same result. In fact you take logs using any base and the answer above remains the same.
Ofc a mathematician would use log_e
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#16

i understand this question up until the point where I'm told to raise both side to the power of three, i put in log3 (-1) and i got a math error im guessing thats why they're doing something. My question is why has log3 disappeared and -1 has changed to 3^-1 and not -1^3
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6 months ago
#17
(Original post by A0W0N)

i understand this question up until the point where I'm told to raise both side to the power of three, i put in log3 (-1) and i got a math error im guessing thats why they're doing something. My question is why has log3 disappeared and -1 has changed to 3^-1 and not -1^3
Inverse logs is the same as raising the number to the base. The log is base 3 here.

So raise both sides with a base of 3. The right hand side becomes 3^(-1). The left becomes
3^log(...)
As the log is base 3, this just cancels, leaving the argument which is the term in the brackets.
Last edited by mqb2766; 6 months ago
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#18
(Original post by mqb2766)
Inverse logs is the same as raising the number to the base. The log is base 3 here.

So raise both sides with a base of 3. The right hand side becomes 3^(-1). The left becomes
3^log(...)
As the log is base 3, this just cancels, leaving the argument which is the term in the brackets.
raising which number to the base?
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6 months ago
#19
(Original post by A0W0N)
raising which number to the base?
You're raising the expression on each side of the equation. So
3^log(x^2/(5x+2)) = 3^(-1)
By definition (of a log base 3), the left hand side is just x^2/(5x+2)
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#20
(Original post by mqb2766)
You're raising the expression on each side of the equation. So
3^log(x^2/(5x+2)) = 3^(-1)
By definition (of a log base 3), the left hand side is just x^2/(5x+2)
I'm really sorry i think im just a slow learner but i still don't understand it, the way i see it is (-1)^3 and im guessing it putting it to the power of 3 on the left side doesn't affect (x^2/5x+2) and it just cancels out log 3.
Is there i video i could watch, and what would i search up if so and how would i enter log3 to the power of three like this:
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