The table shows approximate values of x and y that satisfy a relationship of the form y = ab^x. Plot 10(y) against x to verify.

Take the logarithm base 10 of both sides:

$$\log_{10}(y) = \log_{10}(ab^x)$$

Using logarithm properties, $\log(MN) = \log M + \log N$ and $\log(M^p) = p \log M$:

$$\log_{10}(y) = \log_{10}(a) + \log_{10}(b^x)$$ $$\log_{10}(y) = \log_{10}(a) + x \log_{10}(b)$$

This equation is in the form of a straight line $Y = mX + C$, where:
$Y = \log_{10}(y)$
$X = x$
$m = \log_{10}(b)$ (gradient)
$C = \log_{10}(a)$ (Y-intercept)

Calculate $\log_{10}(y)$ for each given $y$ value (rounded to 3 decimal places for plotting accuracy):

$X = x$	$Y = \log_{10}(y)$
10	$\log_{10}(3.47) \approx 0.540$
20	$\log_{10}(5.50) \approx 0.740$
35	$\log_{10}(10.47) \approx 1.020$
50	$\log_{10}(20.89) \approx 1.320$
60	$\log_{10}(33.11) \approx 1.520$
80	$\log_{10}(79.43) \approx 1.900$

To verify the relationship $y = ab^x$, plot the points $(X, Y)$ from the table above, i.e., plot $\log_{10}(y)$ on the vertical axis against $x$ on the horizontal axis.

• Choose appropriate scales for both axes to accommodate the range of values.
• Plot the six points: $(10, 0.540)$, $(20, 0.740)$, $(35, 1.020)$, $(50, 1.320)$, $(60, 1.520)$, $(80, 1.900)$.
• Draw a line of best fit through these plotted points.
• If the points lie approximately on a straight line, then the relationship $y = ab^x$ is verified. Observing the calculated points, they appear to form a nearly straight line.

From the line of best fit drawn in Step 3, determine its gradient ($m$) and Y-intercept ($C$).
Let's use two points from the line of best fit to calculate the gradient and Y-intercept. For demonstration, we'll use the first and last transformed data points as if they were read from the line of best fit: $(X_1, Y_1) = (10, 0.540)$ and $(X_2, Y_2) = (80, 1.900)$.

Calculate the gradient $m$:

$$m = \frac{Y_2 - Y_1}{X_2 - X_1} = \frac{1.900 - 0.540}{80 - 10} = \frac{1.360}{70} \approx 0.019428$$

Calculate the Y-intercept $C$ using the equation $Y = mX + C$ and one of the points, e.g., $(10, 0.540)$:

$$0.540 = (0.019428)(10) + C$$ $$0.540 = 0.19428 + C$$ $$C = 0.540 - 0.19428 = 0.34572$$

Now, relate $m$ and $C$ back to $a$ and $b$:

$$m = \log_{10}(b) \implies b = 10^m$$ $$C = \log_{10}(a) \implies a = 10^C$$

Substitute the calculated values for $m$ and $C$:

$$b = 10^{0.019428} \approx 1.0456$$ $$a = 10^{0.34572} \approx 2.2166$$

Rounding the answers to 2 decimal places:

$$a \approx 2.22$$ $$b \approx 1.05$$

The approximate values of the constants are $a = 2.22, b = 1.05$.