Skill: Interpret parameters in exponential and logarithmic models

To interpret a model, identify what each parameter does and look at the value when the input is $0$ (for an exponential model) or where the logarithm’s input is $0$ (for a logarithmic model).

Exponential models

For

$$y=ab^x,$$

• $a$ is the initial value because $y=a$ when $x=0$.
• $b$ is the multiplicative factor for each increase of $1$ in $x$.
  • If $b>1$, the model shows growth.
  • If $0<b<1$, the model shows decay.

For

$$y=Ae^{kx},$$

• $A$ is the initial value.
• $k$ is the continuous growth or decay rate.
  • If $k>0$, there is continuous growth.
  • If $k<0$, there is continuous decay.
• The factor for one unit of increase in $x$ is $e^k$.

Logarithmic models

For

$$y=a+b\log_c(x-h),$$

• $a$ shifts the graph vertically.
• $b$ vertically stretches or compresses the graph. If $b<0$, it also reflects the graph.
• $h$ shifts the graph horizontally.
• The domain is $x>h$ because the logarithm’s input must be positive.
• The vertical asymptote is $x=h$.
• A logarithm represents additive changes in the output, not constant multiplicative changes.

Worked example

Interpret the parameters in the model

$$y=4-2\log_3(x-5).$$

Step 1: Identify the vertical shift.

The number $4$ is added outside the logarithm, so the graph is shifted up $4$ units.

Step 2: Identify the vertical scaling and reflection.

The coefficient of the logarithm is $-2$.

• The absolute value $2$ means the graph is vertically stretched by a factor of $2$.
• The negative sign reflects the graph across the $x$-axis.

Step 3: Identify the horizontal shift.

The expression is $x-5$, so the graph shifts right $5$ units.

Step 4: Find the domain and vertical asymptote.

The logarithm’s input must be positive:

$$x-5>0,$$

so

$$x>5.$$

Therefore, the domain is $x>5$, and the vertical asymptote is

$$x=5.$$

So, $y=4-2\log_3(x-5)$ is a logarithmic graph shifted right $5$ units and up $4$ units, vertically stretched by $2$, and reflected across the $x$-axis.