solve
4^3x-4=32^x+2
Please help!
Do you mean
4^(3x) -4 = 32^x +2 ?
yes
If that is really what you mean, I cannot get a closed-form solution for you. You would have to solve it by trial-and-error or graphical means. If, on the other hand, you mean
4^(3x-4)=32^(x+2), then take logs of both sides and get
(3x-4)log 4 = (x+2) log 32
[(3x-4)/(x+2)] = log [4^(5/2)]/log 4
[(3x-4)/(x+2)] = 2.5
3x - 4 = 2.5 x + 5
0.5 x = 9
x = 18
To solve the equation 4^(3x - 4) = 32^(x + 2), we can take logs of both sides.
Using the property log(a^b) = b * log(a), we can rewrite the equation as:
(3x - 4) * log(4) = (x + 2) * log(32)
Since log(4) = log(2^2) = 2 * log(2), and log(32) = log(2^5) = 5 * log(2), we have:
(3x - 4) * 2 * log(2) = (x + 2) * 5 * log(2)
Simplifying further, we get:
2(3x - 4) = 5(x + 2)
Multiply through:
6x - 8 = 5x + 10
Subtract 5x from both sides and add 8 to both sides:
6x - 5x = 10 + 8
x = 18
Therefore, the solution to the equation is x = 18.