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, 201411 yr by Function

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

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.