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Write an exponential function in the form 
y=ab^(x) that goes through the points 
(0,19) and 
(7,2432).
Answer:

Write an exponential function in the form y=abx y=a b^{x} that goes through the points (0,19) (0,19) and (7,2432) (7,2432) .\newlineAnswer:

Full solution

Q. Write an exponential function in the form y=abx y=a b^{x} that goes through the points (0,19) (0,19) and (7,2432) (7,2432) .\newlineAnswer:
  1. Find 'a' value: To find the exponential function, we need to determine the values of 'a' and 'b' in the equation y=abxy=ab^{x}. We can use the point (0,19)(0,19) to find 'a' because any number raised to the power of 00 is 11. So, the equation becomes y=a1y=a\cdot 1, which simplifies to y=ay=a. Since the function passes through (0,19)(0,19), we can set yy to 1919 and solve for 'a'.\newlineCalculation: 19=a119 = a \cdot 1\newline(0,19)(0,19)00
  2. Find 'b' value: Now that we have the value of 'a', we need to find 'b'. We can use the second point (7,2432)(7,2432) to find 'b'. We substitute 'a' with 1919 in the equation y=abxy=ab^{x} and then solve for 'b'.\newlineCalculation: 2432=19b72432 = 19 \cdot b^{7}\newlineb7=243219b^{7} = \frac{2432}{19}\newlineb7=128b^{7} = 128
  3. Calculate 'b': To find 'b', we need to take the seventh root of 128128. The seventh root of a number is the same as raising that number to the power of 17\frac{1}{7}.\newlineCalculation: b=(128)17b = (128)^{\frac{1}{7}}\newlineb2b \approx 2
  4. Check approximation: We need to check if b2b \approx 2 is a correct approximation. Since 27=1282^7 = 128, we can confirm that b=2b = 2 is the exact value and not just an approximation.\newlineCalculation: 27=1282^7 = 128
  5. Write exponential function: Now that we have both aa and bb, we can write the exponential function. We have a=19a = 19 and b=2b = 2, so the function is y=19×2xy = 19 \times 2^{x}.

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