8^(2-x)=4^(3x) x=? I need to know how to get to the answer, thanks.?

Answer 1

First express both 8 and 4 in terms of powers of 2. (8 = 2^3 and 4 = 2^2.) That gives one:

2^[3(2 - x)] = 2^[2(3x)], so that equating the exponents,

6 - 3x = 6x. Thus 6 = 9x, and therefore

x = 2/3.

CHECK that x = 2/3 solves the ORIGINAL equation:

With x = 2/3, 2 - x = 2 - 2/3 = 4/3, and so the LHS, 8^4/3 = 2^4 = 16.

Meanwhile, on the RHS 3x = 2, and 4^2 = 16.

So both sides of the original equation are equal. Thus the result x = 2/3 is CORRECT.

QED

Live long and prosper.

Answer 2

first you have to make the bases the same so,
(2^3)^2-x=(2^2)^3x

Now multiply the exponents:
2^6-3x=2^6x

6-3x=6x
6=3x
divide each side my 3 so your left with x
2=x

Answer 3

8^(2-x) =4^(3x)

8 = 2^3 and 4=2^2 so

8^(2-x) = 2^3(2-x) and 4^(3x) = 2^6x
3(2-x) = 6x
6-3x = 6x
9x = 6
x = 2/3

Answer 4

(4*2)^(2-x) = 4^(3x) x
= 4^(2-x) 2^(2-x) = 4^(3x) x
= 4^2 / (4^x) 2^2 / 2^x = 4^3 (4^x) x
= 16 (4) / (2^2x * 2^x)= 64 (2^2x) x
= 1 / (2^2x * 2^x)= 16 (2^2x) x
1 = 16 (2^2x) x [ (2^2x * 2^x) ]
1 = 16 ( 2^5x )
2 = 32 ( 2^5x )
2 = 2^(5x+5) . . . .. consider the exponential equation
1 = 5x + 5
x = - 4/5

Answer 5

8^(2-x)=4^(3x)
4^2(2-x)=4^(3x)
4-4x=3x
4=7x
x=4/7