Recall that the slope of the tangent line at a point is given by the derivative at that $x$-value. You need $g'(x)$ and specifically $g'(1)$.

For a constant base $c$, the derivative of $c^{x}$ is not a power rule; it is $c^{x} \ln c$. Use this to write $g'(x)$ and set $g'(1)$ equal to the given slope $8\ln 2$.

Once you have an equation like $c \ln c = 8 \ln 2$, use log properties (such as $a\ln b = \ln(b^{a})$) to rewrite it in a more recognizable exponential form, then look for a simple value of $c$ that works.

In Desmos, enter the function f(x) = x*ln(x) to represent the left-hand side $x \ln x$.

Enter a second expression g(x) = 8*ln(2); this will appear as a horizontal line at the height $y = 8\ln 2$.

Use the intersection tool (tap on the point where the graphs meet) to find the x-coordinate where f(x) and g(x) intersect; that x-value is the value of $c$ that satisfies $c \ln c = 8 \ln 2$.

The function is defined as $g(x)=c^{x}+d$. Since the graph passes through $(1,8)$, plug in $x=1$ and $g(1)=8$:

So we know

We will use this later only to check our work; it is not needed to find $c$ directly.

To use the slope of the tangent line, we need the derivative of $g$.

For an exponential function with constant base $c$, the derivative of $c^{x}$ is $c^{x} \ln c$. So

The tangent slope at $x=1$ is $g'(1)$, which is given as $8\ln 2$:

Now we have an equation involving only $c$:

Use the property $a \ln b = \ln(b^{a})$.

Apply it to both sides of $c \ln c = 8 \ln 2$:

• Left side: $c \ln c = \ln(c^{c})$.
• Right side: $8 \ln 2 = \ln(2^{8}) = \ln 256$.

So the equation becomes

Because the natural log function is one-to-one (if $\ln A = \ln B$, then $A=B$), we get

We look for a positive number $c$ such that $c^{c} = 256$. Try small integer values:

• $2^{2} = 4$
• $3^{3} = 27$
• $4^{4} = 256$

This matches the required value, so $c = 4$.

Therefore, the value of $c$ is 4.