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Maths logs

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

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Reply 1
Avatar for A0W0N
A0W0N
OP
9
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6i)
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you have to use laws of logs. log(an)=nloga
Remember the laws of logs:
Unparseable latex formula:

\log\left(a\right)+\log\left(b\)=\log\left(ab\right)


Unparseable latex formula:

\log\left(a\right)-\log\left(b\)=\log\left(\frac{a}{b}\right)


blog⁡(a)=log⁡(ab)b\log\left(a\right)=\log\left(a^{b}\right)
Reply 4
Avatar for A0W0N
A0W0N
OP
9
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thanks i got that im stuck on this one now
Reply 5
Avatar for mqb2766
mqb2766
21
Original post by A0W0N
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thanks i got that im stuck on this one now

Which one?
Reply 6
Avatar for A0W0N
A0W0N
OP
9
7ii)
Reply 7
Avatar for A0W0N
A0W0N
OP
9
image.jpg
i don't don't how to find x
Original post by A0W0N
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i don't don't how to find x

you can remove logs from both sides because you know n0 = 1, where n =/= 0
Reply 9
Avatar for mqb2766
mqb2766
21
Original post by A0W0N
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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
Reply 10
Avatar for A0W0N
A0W0N
OP
9
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why is it telling me to use base 10?
(i put what i think the answer is on the calculator)
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Original post by A0W0N
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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.
Reply 12
Avatar for A0W0N
A0W0N
OP
9
Original post by mqb2766
The right hand side are (mostly) easy to calculate using base 10.

but wouldn't they produce different results?
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
..
(edited 4 years ago)
Reply 14
Avatar for B_9710
B_9710
18
Original post by A0W0N
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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
Reply 15
Avatar for A0W0N
A0W0N
OP
9
image.jpg
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
Original post by A0W0N
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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.
(edited 4 years ago)
Reply 17
Avatar for A0W0N
A0W0N
OP
9
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?
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)
Reply 19
Avatar for A0W0N
A0W0N
OP
9
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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