3^n-1=2^n+1
how can i solve this problems, please help?
nasty and tricky the way you typed it
did you mean:
3^(n-1) = 2^(n+1) ??
I sure hope so, and will assume that.
take log of both sides
log [3^(n-1) ] = log [2^(n+1) ]
(n-1) log3 = (n+1) log2
n log3 - log3 = n log2 + log2
n log3 - n log2 = log2 + log3
n(log3 - log2) = log2 + log3
n = (log2+log3)/(log3-log2)
= appr 4.42
it verifies in the original.
I assume it is really this:
3^(n-1)=2^(n+1)
take the log of each side
(n-1)log3=(n+1)log2
now solve for n.
To solve this equation, we need to find the value of n that satisfies the equation 3^n-1=2^n+1. Here's how you can solve it:
1. Start by rearranging the equation to isolate the terms with n on one side. Add 1 to both sides of the equation to move -1 to the right side:
3^n - 1 + 1 = 2^n + 1 + 1
Simplifying this gives us:
3^n = 2^n + 2
2. Now, let's subtract 2^n from both sides to isolate the 2 on the right side:
3^n - 2^n = 2
3. Next, we can factor out 2^n from the left side of the equation:
2^n * (3^n/2^n - 1) = 2
4. Now, divide both sides of the equation by 2^n:
3^n/2^n - 1 = 2^(1-n)
5. We can simplify further by using the fact that 2^(1-n) is equal to (2^n)/(2^n):
3^n/2^n - 1 = (2^n)/(2^n)
6. Combine the fractions on the left side:
(3^n - 2^n)/2^n = 1
7. Multiply both sides of the equation by 2^n to remove the fraction:
3^n - 2^n = 2^n
8. Now, we have a simpler equation:
3^n = 3 * 2^n
9. Divide both sides of the equation by 3:
3^(n-1) = 2^n
10. Now, we have two sides with the same base (3) raised to different exponents. The only way for the equation to hold true is if the exponents are equal:
n - 1 = n
11. Subtract n from both sides:
-1 = 0
12. We have arrived at a contradiction. This means that there is no value of n that satisfies the original equation 3^n - 1 = 2^n + 1. Therefore, the equation has no solution.
In summary, by simplifying and manipulating the equation, we found that there is no value of n that satisfies 3^n - 1 = 2^n + 1.