Question

How do you find the constants a and b?

`f(x) = aln(bx)`

where `f(e)` and `f'(2) = 2`

Answer 1

`a\ =\ 4`
`b\ =\ \frac{1}{\sqrt{e}}`

Explanation:

The first step is to find the derivative of the given function and then put the value of `x=2` in the derivative function. The solve the equation for the variable left.

`f^{'}(x)\ =\ \frac{d}{dx}(a\ln (bx))`

`=a\frac{d}{dx}(\ln (bx))`

Apply the derivative rule `\frac{d}{du}(\ln (u))=\frac{1}{u}` to get:

`=a\cdot \frac{1}{bx}\cdot b`

Simplify:

`f^{'}(x)\ =\ \frac{a}{x}`

Put `f^{'}(x)=2` to get:

`f^{'}(2)\ =\ \frac{a}{2}`

Since, `f^{'}(2)\ =\ 2`, by putting here, we get:

`2=a/2`

Simplify:

`a=4`

Now, by putting `x=e` in the original function, we have:

`f(e)=4\cdot ln\(b\cdot e)`

To solve for `b` , put the value of `a` and `f(e)=2` to get:

`2=4\cdot ln\(b\cdot e)`

Apply the lograithm product rule to rewrite it as:

`ln(m\cdot n)\ =\ ln(m)+ln(n)`

`2=4[ln(b)+ln(e)]`

By manipulating the sides, you will have:

`\ln (b)=\frac{1}{2}-\ln (e)`

Put `ln(e)=1`

`\ln (b)=\frac{1}{2}-1`

Solve for `b` to get:

`b=\frac{1}{\sqrt{e}}`

That's it!