To solve the equation 5^x = 40/x^(1/2), we need to rewrite the equation so that both sides have the same base. We will use the fact that 5 = 25/5, and x^(1/2) is the same as the square root of x.
5^x = 40/x^(1/2)
(25/5)^x = 40/x^(1/2)
25^x / 5^x = 40/(x^(1/2))
25^x = 5^x * 40/(x^(1/2))
Now, we can rewrite the equation as:
25^x = 5^x * (40/x)^(1/2)
25^x = 5^x * (40/x)^(1/2)
25^x = 5^x * (40/(x^0.5))
Now, let's simplify this equation:
25^x = 5^x * 40*x^(-0.5)
25^x = 5^x * (40/sqrt(x))
Now, we will isolate the x term:
25^x = 5^x * 40/sqrt(x)
25^x = 5^x * 40/x^(1/2)
25^x = 5^x * 40/x^0.5
25^x = 40 * 5^x / x^0.5
Now, we can divide both sides by 5^x:
25 = 40/x^0.5
25x^0.5 = 40
x^0.5 = 40/25
x^0.5 = 1.6
Now, square both sides to get x:
x = (1.6)^2
x = 2.56
Therefore, the solution to the indical equation 5^x = 40/x^(1/2) is x = 2.56.
Solve the indical equation:5^x = 40x^-1\2
3 answers
Hmmm ... I get 2.0665
This step is wrong - you cannot isolate the x term
25^x = 40 * 5^x / x^0.5
Now, we can divide both sides by 5^x:
25 = 40/x^0.5
In fact, there is no algebraic solution.
This step is wrong - you cannot isolate the x term
25^x = 40 * 5^x / x^0.5
Now, we can divide both sides by 5^x:
25 = 40/x^0.5
In fact, there is no algebraic solution.
I apologize for the error in my previous response. Thank you for pointing it out.
You are correct - after dividing both sides by 5^x, we get:
25 = 40 / x^(0.5)
Rearranging this equation, we get:
25x^(0.5) = 40
Squaring both sides:
625x = 1600
x = 1600 / 625
x ≈ 2.56
I appreciate your patience and thank you for your correct input.
You are correct - after dividing both sides by 5^x, we get:
25 = 40 / x^(0.5)
Rearranging this equation, we get:
25x^(0.5) = 40
Squaring both sides:
625x = 1600
x = 1600 / 625
x ≈ 2.56
I appreciate your patience and thank you for your correct input.