Solve: (x-2)5^x = 5^(x+2)
(x-2)5^x - 5^(x+2) = 0
I see a common factor of 5^x
5^x( x-2 + 5^2) = 0
5^x(x + 23) = 0
5^x = 0 or x = -23
but 5^x = 0 has no solution, so
x = -23
take the log base 5 of each side.
log(x-2)+x =(x+2)
log(x-2)=2
log base five z=2, so x-2=25, x= 27
check it.
x-2)5^x = 5^(x+2)
25*5^25= 5^(27)
25= 5^2, checks.
go with bobpursley's solution
I don't know how my "-" suddenly turned into a + in my third line.
should have been:
5^x( x-2 - 5^2) = 0
5^x(x - 27) = 0
5^x = 0 or x = 27
but 5^x = 0 has no solution, so
x = 27 , the same as bob had
To solve the equation (x-2)5^x = 5^(x+2), we need to use logarithms and apply some algebraic manipulation.
Step 1: Start by dividing both sides of the equation by 5^x to eliminate the exponential terms:
(x - 2) = 5^2
Step 2: Simplify the equation:
x - 2 = 25
Step 3: Add 2 to both sides:
x = 27
Therefore, the solution to the equation (x-2)5^x = 5^(x+2) is x = 27.