# Finding a constant

1. Dec 2, 2005

### Werg22

Is it possible to find a constant value for b in the following equality for any value of x?

$$b^{x}(\ln b) = a^x$$

2. Dec 2, 2005

### BerkMath

what is a? Is it given? If so, in what field is it an element.

3. Dec 2, 2005

### Werg22

a can be any value, so the real question is to find b in relationship to a, independantly of x.

Last edited: Dec 2, 2005
4. Dec 2, 2005

### HallsofIvy

Staff Emeritus
Since you didn't answer the question "in what field is it an element?" I assume this is in the real numbers.

$$b^{x}(\ln b) = a^x$$
Taking ln of both sides:
$$ln(b^x ln b)= ln b^x + ln b= xln b+ ln b= x ln a$$
so
$$x(ln a- ln b)= x ln\frac{a}{b}= ln b$$
That will be "independent of x" if and only if
$$ln \frac{a}{b}= 0$$
or
a= b= 1.

5. Dec 2, 2005

### Werg22

I think you made a mistake

You're development:

$$\ln(b^x \ln b)= \ln b^x + \ln b$$

When it should be

$$\ln(b^x \ln b)= \ln b^x + \ln(\ln b)$$

Last edited: Dec 2, 2005
6. Dec 2, 2005

### CRGreathouse

Suppose $$a,b\in\mathbb{R}^+$$. If $$b^x\ln b=a^x$$ then we have
$$x\ln b+\ln\ln b=x\ln a$$
or
$$\ln\ln b=x(\ln a-\ln b)$$.

If $$a\neq b$$ then this will not hold for all $$x\in\mathbb{R}$$. Then we have a=b and $$\ln\ln b=0$$. This means the unique solution is $$a=b=e$$.

7. Dec 2, 2005

### Werg22

Does that mean that the function a^x has no integral function if a is not equal to e?

8. Dec 3, 2005

### HallsofIvy

Staff Emeritus
???
$$a^x= e^{ln a^x}= e^{x ln a}$$
so the anti-derivative (I guess that's what you mean by "integral function") is
[tex]\frac{e^{x ln a}}{ln a}= \frac{a^x}{ln a}[/itex].

Yes, werg22. Thanks for the correction.