how would I solve 8^(2x+3) = 4 * 2^(x+1)?

Answer 1

8^(2x+3) = 4 * 2^(x+1)

2^3(2x+3)=2^2 *2^(x+1)
2^3(2x+3)=2^(2+x+1)
since the bases are equal
3(2x+3)=2+x+1
6x+9=3+x
5x=-6
x=-6/5

Answer 2

Very carefully.
Recognize that 8=2^3 and 4=2^2
Then, putting everything into base 2
2^(3*[2x+3])= 2^(x+3)
Now you can set the exponents equal to each other and solve for x

Answer 3

8^(2x + 3) = 4 * 2^(x + 1)

2^(3 * (2x + 3)) = 2^2 * 2^(x + 1)

2^(6x + 9) = 2^(2 + x + 1)

2^(6x + 9) = 2^(x + 3)

so;

6x + 9 = x + 3

5x = -6

x = -6/5 = -1.2

Answer 4

8^(2x+3) = (2^3)^(2x+3) = 2^(6x+9)

4*2^(x+1) = 2^(x+3)

so equating both sides:

2^(6x+9) = 2^(x+3)
and multiplying both sides by 2^(-(x+3)) to get 1 on the RHS:
2^(6x+9)*2^(-(x+3)) = 2^(5x+6) = 1

take the log(base2) of each side:

5x+6 = 0
x = -6/5...

Hope that helps...

Answer 5

log(8^(2x + 3)) = log(4*2^(x +1))
(2x+3)*log8 = log4 + (x +1)*log2
2x*log8 + 3*log8 = log4 + x*log2 + log2
2x*log(2^3) + 3*log(2^3) = log(2^2) + x*log2 + log2
6x*log2 + 9*log2 = 2*log2 + x*log2 + log2

eliminate (log2), it will be

6x + 9 = 2 + x + 1
5x = -6
x = -(6/5)