7^x + 6^x - 2^x = 9^x
List all integer solutions of the above equation.
Lets put some bounds on the possible values of x.
For positive x then 9^x>7^x>6^x>2^x.
Then we can create a compound inequality:
9^x < 9^2 + 2^x = 7^x + 6^x < 2*7^x
Take the extreme left and right and solve for x:
9^x < 2*7^x
x * log 9 < log 2 + x * log 7
x < log 2 / (log 9 - log 7) = 2.758
For negative x then 2^x>6^x>7^x>9^x
Then this compound inequality can be formed:
2^x < 9^2 + 2^x = 7^x + 6^x < 2*6^x
Again,take the extreme left and right and solve for x:
2^x < 2*6^x
x * log 2 < log 2 + x * log 6
x > log 2 / (log 2 - log 6) = -0.631
Then x must be in the interval (-0.631, 2.758). Since we are asked for only integer solutions then trial testing integers {0,1,2} is all that is needed since those are the only integers in the interval.
7^0 + 6^0 - 2^0 = 9^0 simplifies to 1=1; therefore x=0 is a solution.
7^1 + 6^1 - 2^1 = 9^1 simplifies to 3=9; therefore x=1 is not a solution.
7^2 + 6^2 - 2^2 = 9^2 simplifies to 81=81; therefore x=2 is a solution.
Then the list of all integer solutions to the given equation is x=0 and x=2.
Solution
