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#1 2023-12-17 08:42:39
- paulb203
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- Registered: 2023-02-24
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What is the method for; 3^a = 1/9 Solve for a. ?
I was able to solve this but through trial and error rather than method; what is the method?
P.S.
My trial and error went something like;
3^0=1
3^1=3
3^2=9
3^3=27
This is only producing integers so try something else;
3^-1 = 1/3
3^-2 = 1/9
Ah, so n = -2
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#2 2023-12-17 14:16:45
- KerimF
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Re: What is the method for; 3^a = 1/9 Solve for a. ?
Perhaps you mean the following method:
3^a = 1/9
3^a = 1/(3^2)
3^a = 3^(-2)
a= -2
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But only a human may have the freedom and ability to oppose his natural robotic nature.
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#3 2023-12-17 14:41:26
- Jai Ganesh
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Re: What is the method for; 3^a = 1/9 Solve for a. ?
Hi,
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#4 2023-12-18 05:48:51
- paulb203
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- Registered: 2023-02-24
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Re: What is the method for; 3^a = 1/9 Solve for a. ?
Perhaps you mean the following method:
3^a = 1/9
3^a = 1/(3^2)
3^a = 3^(-2)a= -2
Ah, thanks. I love it.
So, for a further example
4^n = 1/16
4^n = 1/4^2
4^n = 4^-2
n=-2
And,
2^x = 1/4
2^x = 1/2^2
2^x = 2^-2
x=-2
Q. But presumably this only works if the numerator of the fraction on the right is a square of the integer value on the left (9 is a square of 3; 16 is a square of 4; and 4 is a square of 2)?
Prioritise. Persevere. No pain, no gain.
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#5 2023-12-18 05:50:06
- paulb203
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Re: What is the method for; 3^a = 1/9 Solve for a. ?
Thanks, Jai Ganesh
Prioritise. Persevere. No pain, no gain.
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#6 2023-12-18 09:28:40
- amnkb
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- Registered: 2023-09-19
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Re: What is the method for; 3^a = 1/9 Solve for a. ?
Q. But presumably this only works if the numerator of the fraction on the right is a square of the integer value on the left (9 is a square of 3; 16 is a square of 4; and 4 is a square of 2)?
Not necessarily
You could have something like 4^n=8
so switch to (2^2)^n = 2^(2n) and 8 = 2^3
2n=3
n=3/2
But yeh it has to be the same base somehow (or else logs, i think)
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#7 2023-12-19 08:59:11
- paulb203
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Re: What is the method for; 3^a = 1/9 Solve for a. ?
paulb203 wrote:Q. But presumably this only works if the numerator of the fraction on the right is a square of the integer value on the left (9 is a square of 3; 16 is a square of 4; and 4 is a square of 2)?
Not necessarily
You could have something like 4^n=8
so switch to (2^2)^n = 2^(2n) and 8 = 2^3
2n=3
n=3/2
But yeh it has to be the same base somehow (or else logs, i think)
Thanks, amnkb.
I thought I wasn't going to be able to follow that but managed with a bit of work. Cheers.
Prioritise. Persevere. No pain, no gain.
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