Given that 4^X=8X. Solve for X
tough one!
are you sure you did not mean
4^x = 8^x ?
that would be easy.
What you typed requires some sort of sophisticated method such as "Newton's Method"
but an intuitively obvious answer would be x = 2
however there is another one between 0 and 1
let me know which way it is
Yeah X=2. I just found an answer after 7 hours of solving. You don't need newton's method here. All you need is to apply that basic logarithm rules
To solve the equation 4^X = 8X, we need to find the value of X that satisfies the equation.
To start, we can rewrite 8X as 2^3X since 8 is equal to 2 cubed.
So the equation becomes: 4^X = 2^3X
Now, let's take the logarithm of both sides of the equation to get rid of the exponential terms. We can use any base for the logarithm, but let's use the common logarithm (base 10) for this example:
log(4^X) = log(2^3X)
Using the power rule of logarithms, we can rewrite the equation as:
X * log(4) = 3X * log(2)
Next, simplify by dividing both sides of the equation by X:
log(4) = 3 * log(2)
Now, we can solve for log(2) by dividing both sides of the equation by 3:
log(2) = log(4) / 3
Using the fact that log(a^b) = b * log(a), we can rewrite the equation as:
log(2) = log(2^2) / 3
log(2) = 2 * log(2) / 3
Next, multiply both sides of the equation by 3:
3 * log(2) = 2 * log(2)
Now, subtract 2 * log(2) from both sides of the equation:
3 * log(2) - 2 * log(2) = 0
log(2) = 0
Finally, solve for X by substituting the value of log(2) back into the equation:
X = log(2) / log(4)
Using a calculator, we can evaluate log(2) and log(4):
X ≈ 0.693 / 0.602
X ≈ 1.151
Therefore, the approximate solution to the equation 4^X = 8X is X ≈ 1.151.