Math Is Fun Forum

paulb203

Member

Registered: 2023-02-24

Posts: 136

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

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 164

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

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 46,281

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.

paulb203

Member

Registered: 2023-02-24

Posts: 136

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)?

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: What is the method for; 3^a = 1/9 Solve for a. ?

Thanks, Jai Ganesh

amnkb

Member

Registered: 2023-09-19

Posts: 253

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)

paulb203

Member

Registered: 2023-02-24

Posts: 136

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)

Thanks, amnkb.
I thought I wasn't going to be able to follow that but managed with a bit of work. Cheers.