'Special' logarithmic function

[Function]

Hi everyone

 

Out of boredom, I asked myself the following question:

 

For which [math]a[/math], [math]b[/math], [math]c[/math], [math]d[/math], [math]e[/math] and [math]f[/math] does the curve, described by the function

 

[math]g(x)=a\cdot\log_b{\left(c\cdot x^d+e\right)}+f[/math]

 

Only have one solution, so touches the curve described by the function

 

[math]h(x)=x[/math]

 

In [math](1,1)[/math]?

 

Found something, and wanted to know if it was right. It's been a while since I 'performed' pure mathematics (over 4 months), so don't be too hard on me

 

The derivative of that function in [math]x=1[/math] should be 1.

 

[math]\frac{d}{dx}a\cdot\log_b{\left(c\cdot x^d+e\right)}+f = 1[/math]

 

[math]\Leftrightarrow a\cdot\frac{d}{dx}\left(\log_b{\left(c\cdot x^d+e\right)}\right) = 1[/math]

 

[math]\Leftrightarrow a\cdot \frac{c\cdot d\cdot x^{d-1}}{\left(c\cdot x^d+e\right)\cdot\ln{b}} = 1[/math]

 

(If there's a mistake, it's in the equation above this line... forgot how the derivative of a logarithmic function, so had to do it manually and found that the derivative of y to x of the function y = log_b{x} is 1/(xlnb))

 

[math]\Leftrightarrow a\cdot c\cdot d\cdot x^{d-1} = \left(c\cdot x^d+e\right)\cdot\ln{b}[/math]

 

[math]x = 1[/math]

 

[math]\Leftrightarrow a\cdot c\cdot d = (c+e)\cdot\ln{b}[/math]

 

Second possible expression:

 

[math]a\cdot\log_b{\left(c+e\right)}+f=1[/math]

 

Are these valid expressions to get the wanted?

 

Thanks.

 

F.

Edited November 24, 2014 by Function

[elfmotat]

I'm not sure what you mean by "only have one function, so touches the curve described by the function." Could you explain?

[Function]

I'm not sure what you mean by "only have one function, so touches the curve described by the function." Could you explain?

 

replaced 'function' with 'solution'

so that the tangent in x = 1 of that function is the function x.