The given problem involves an exponential function of the form:

$y = a \times b^x$

From the graph, we have two points: $(2, 27)$ and $(3, 54)$.

We can substitute these points into the exponential equation to form two equations:

1. $27 = a \times b^2$
2. $54 = a \times b^3$

Step 1: Divide the equations to eliminate $a$.

Dividing equation (2) by equation (1):

$\frac{54}{27} = \frac{a \times b^3}{a \times b^2}$

This simplifies to:

$2 = b$

Step 2: Solve for $a$.

Now substitute $b = 2$ into one of the original equations. Let's use equation (1):

$27 = a \times 2^2$

This simplifies to:

$27 = a \times 4$

Solving for $a$:

$a = \frac{27}{4} = 6.75$

Final Answer:

$a = 6.75, \quad b = 2$

Would you like a further explanation or have any questions?

Here are 5 related questions to extend your learning:

1. What is the significance of the base $b$ in an exponential function?
2. How would the graph change if the base $b$ was less than 1?
3. Can an exponential function have a negative base?
4. What is the effect of changing the value of $a$ in an exponential equation?
5. How can you use logarithms to solve exponential equations more generally?

Tip: For exponential functions, the value of $b$ determines the rate of growth or decay. If $b > 1$, the function grows; if $0 < b < 1$, the function decays.