3^x+4=9^x-1 can you please help me thank you

The answer is x = 0.93..

Step-by-step explanation:

- I converted 9^x to base 3

-  Then I applied the exponent rule( a^b)^c = a ^bc

- After that I applied the other exponent rule a^bc = ( a^b) ^c

- I then rewrited the equation and solved it

- And I got x= 0.93..

Please BRAINLIEST!

you divide them and then you get the answer i took this class this year

step-by-step explanation:

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: