- Download PDF
2 Answers | Add Yours
To solve for the equation :
5^x-6^x = 3^x-4^x.
We can see that x = 0 gives:
LHS 5^0-6^0 = 1-1 = 0.
RHS = 3^0-4^0 =0
So the equation satisfies for x = 0.
Similarly for x =1 ,
LHS: 5^1-6^1 = 5-6 = 0.
RHS: 3^1-4^1 = 1-1 = 0.
Therefore x = 0 and x =1 are the two roots of the given exponential equation.
We'll solve the equation using Lagrange's theorem on the following intervals, [3,4] and [5,6] for the function f(x):
Before using the theorem, we'll re-write the equality in this manner:
According to Lagrange's theorem, there is a value c, in the interval (5,6) and a value d, in the interval (3,4), so that:
Note that we'll calculate the first derivative having as variable c, respectively d, so, we'll differentiate a power function and not an exponential function.
We'll substitute both sides of equality by (1) and (2)
We'll remove the brackets:
We'll divide by x both sides:
We’ve answered 324,107 questions. We can answer yours, too.Ask a question