How do you find the constants a and b?

#f(x) = aln(bx)#

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

1 Answer
Feb 22, 2018

#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!