The general form of an exponential equation is y=ab^x+k. What is the general form of the following equation?
y−3=4(5^2x+1)
Responses:
y=20(25)x+3y
y=4(25)x+4y
y=4(125)x+3y
y=20(400)x+3
AAAaannndd the bot gets it wrong yet again!
Of course, all the choices are also incorrect
y−3=4(5^2x+1)
y = 4(5^2x+1)+3
= 4(25^x+1)+3
= 4*25^x + 7
unless maybe you meant
y−3=4(5^(2x+1))
y = 4*5^(2x+1)+3
= 4*5*25^x + 3
= 20*25^x + 3
Bruh...what... Thats wrong Bot. Can't rely on you for nothing smh.
OH bet. Thanks man! Much appreciated!
To determine the general form of the given equation, let's analyze it step by step:
The equation given is:
y - 3 = 4(5^(2x + 1))
First, let's simplify the expression inside the parentheses:
2x + 1 = 2(x) + 2(1) = 2x + 2
Now substitute this expression back into the equation:
y - 3 = 4(5^(2x + 2))
Next, let's simplify the expression within the parentheses:
5^(2x + 2) = 5^(2) * 5^(2x) = 25 * (5^2x)
Now substitute this expression back into the equation:
y - 3 = 4(25 * (5^2x))
Next, let's simplify the expression within the parentheses:
25 * (5^2x) = 25 * (25^x) = 25^(x + 1)
Now substitute this expression back into the equation:
y - 3 = 4(25^(x + 1))
Finally, rearrange the equation to match the general form of an exponential equation (y = ab^x + k):
y = 4(25^(x + 1)) + 3
Therefore, the general form of the given equation is:
y = 4(25^(x + 1)) + 3