Help math question (1/9)ª = 3ªֿ¹ what is the value of ª?

Question

cah you sho me the method too thx!

can you show me the method too thx!

Answer 1

9^(-a) = 3^(a-1)
(3^2)^(-a) = 3^(a-1)
3^(-2a) = 3^ (a-1)
-2a = a-1
a= 1/3

Answer 2

First, realize that 3=1/9 ^-1/2 power so rewritten:

(1/9)^a =((1/9)^-1/2)^a-1

raising the power on the right to the power gives:

(1/9)^a =(1/9)^(-a/2+1/2)
since 1/9=1/9, you can equate the 2 exponents

a=-a/2 + 1/2
adding a/2 to both sides:
3/2a = 1/2
divide both side by 3/2
a=1/3

Answer 3

(1/9)^a = 3^(a-1)
[rule: 9 = 3^2] -> [ ] are here in case you need to explain why this or that

(1/3^2)^a = 3^(a-1)
[rule: 1/x = x^(-1)]

((3^2)^-1)^a = 3^(a-1)
[rule: (x^m)^n) = x^(m*n); inside out!]

(3^(-2))^a = 3^(a-1)
[rule: (x^m)^n) = x^(m*n)]

3^(-2a) = 3^(a-1)
[rule: x^a = x^b => a = b; if bases are equal, than the exponents are equal too]

-2a = a -1
-2a - a = -1
-3a = -1 /:(-3)
a = -1/(-3)
a = 1/3

Check for a = 1/3
(1/9)^(1/3) = 3^(1/3 - 1)
(1/9)^(1/3) = 3^(-2/3)
(1/3^2)^(1/3) = 3^(-2/3)
(3^(-2))^(1/3) = 3^(-2/3)
3^(-2/3) = 3^(-2/3) => OK! a = 1/3 and therefore the solution.

Answer 4

Oh wow, now I know I'm rusty. And back in the day, I was very good at this...

(1/9)ª = 3ªֿ¹

1 to the power of a/9 to the power of a

Gosh, sorry...

Answer 5

Easiest way, without logs, is to make that 3 look like a 1/9.

To do that, you'd have to invert it by making the exponent negative, and square the 3, by multiplying the exponent by 2. Sooo..

(1/9)^a = (1/9)^(-2*(a-1))

Then, since the 1/9's are the same, you can set the exponents = to each other to solve.

a = -2(a-1)

Then distribute etc. do the algebra 1 thing.

Answer 6

(1/9)ª = 3ªֿ¹
(1/3^(2)) = 3^(a-1)
3^(-2) = 3^(a-1)
-2 = a – 1
-1 = a

Answer 7

Taking log of both sides

a*log(1/9)= (a-1)log3 but log 1/9 = log 1 -log9 =0- log3^2 =

- 2log3 so -2alog3 = (a-1)log3 -2a= a-1 a=1/3

Answer 8

a is the power. it could be any number. Like 2^3 is (2)(2)(2)=8.

Answer 9

9^(-a) = 3^(a-1)
(3^2)^(-a) = 3^(a-1)
3^(-2a) = 3^ (a-1)
-2a = a-1
a= 1/3

Double check with the calculator:
(1/9)^a = .48074....
3^(a-1) = .48074....

Answer 10

[1]power a={[3]power[a-1]}*[9]power a
1={[3] power[a-1]}*{[3]power 2a}
1=[3]power[a-1+2a]
1=[3]power[3a-1]
log3[1]=3a-1
a={[log3[1]]+1}/3