Question

Let P(x)=ln(4x-5)^k for some positive-valued constant k. If P'(2)=8, what is the value of k?

Answer 1

The expression we have is:

$P(x)=ln(4x-5)^k$

First, we use the following property of natural logarithms:

$\ln A^k=k\ln A$

the exponent is lowered.

We apply this to our expression:

$P(x)=k\ln (4x-5)$

Next, we use the given condition:

$P(2)=8$

What this means is that when x=2, the value of P(x) is 8.

We need to substitute this two values into our expression:

$\begin{gathered} P(x)=k\ln (4x-5) \\ 8=k\ln (4(2)-5) \end{gathered}$

We solve the expression inside the natural logarithm:

$\begin{gathered} 8=k\ln (8-5) \\ 8=k\ln (3) \end{gathered}$

Now, since ln(3)=1.1 we get:

$8=k(1.1)$

We solve for k by dividing both sides by 1.1:

$\begin{gathered} \frac{8}{1.1}=k \\ 7.3=k \end{gathered}$

The value of k is 7.3