What is your way of solving something like this?

Question

3^x 9^2x = 243

I don't see how this is supposed to be done?
Do you have to do something with the 243?
Can anyone use normal people words to do this please?

Answer 1

Convert everything to the same base (preferably the lower):

3^x * 3^(4x) = 3^5. ... note: 9^2 = (3^2)^2 = 3^4
3^(5x) = 3^5
5x = 5 ..... then you consider only the exponents
x=1

Properties of exponents:
(a^n)^m = a^(nm)
a^n * a^m = a^(n+m)



d:

Answer 2

(3^x)(9^2x) = 243
what you can start with is leaving both x's in the exponent as constant, or they do not matter so you have (3)(9^2) = 243
Solve for the 9^2 first and you get (81)(3) which equals 243. So therefore since you left the x's at 1, x = 1.
If x would equal more than 1, you just know that you have (81^x)(3^x) = 243 and you solve by multiplying the values together and you know with exponents you simply multiply them as well so you have (81*3)^ x*x = 243^ (x^2) or in words two hundred and forty three to the x squared = 243, and you then can see that x is still 1. Guess and check...hope that helps.

Answer 3

243=3^5

9=3^2

3^x . 3^4x =3^5

the bases are same add the exponents for left side

3^5x=3^5

5x=5
x=1

Answer 4

243 = 3(81) = 3(9^2) = (3^1)(9^2)

x=2

Answer 5

(3^x)(9^2x)=243

9^(2x)=(3^2)^(2x)=3^(2x(2))=3^(4x)

So we have

(3^x)(3^4x)=243

3^(5x)=243

3^5=543, so 5x=5

x=1

Answer 6

3^x(9^2x)=243
3^x(3^3x)=243
3^(x+3x)=243
3^4x=243
3^4x=3^4

4x=4
x=1