Is there a constant value for b in b^x (ln b) = a^x for any x?

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In summary, the conversation discusses finding a constant value for b in an equality involving b^x and a^x. It is determined that for a and b in the real numbers, the only solution is when a=b=e. It is also noted that the function a^x has an anti-derivative of a^x/ln a.
  • #1
Werg22
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Is it possible to find a constant value for b in the following equality for any value of x?

[tex] b^{x}(\ln b) = a^x [/tex]
 
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  • #2
what is a? Is it given? If so, in what field is it an element.
 
  • #3
a can be any value, so the real question is to find b in relationship to a, independantly of x.
 
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  • #4
Since you didn't answer the question "in what field is it an element?" I assume this is in the real numbers.

[tex] b^{x}(\ln b) = a^x [/tex]
Taking ln of both sides:
[tex]ln(b^x ln b)= ln b^x + ln b= xln b+ ln b= x ln a[/tex]
so
[tex]x(ln a- ln b)= x ln\frac{a}{b}= ln b[/tex]
That will be "independent of x" if and only if
[tex]ln \frac{a}{b}= 0[/tex]
or
a= b= 1.
 
  • #5
I think you made a mistake

You're development:

[tex]\ln(b^x \ln b)= \ln b^x + \ln b[/tex]

When it should be

[tex]\ln(b^x \ln b)= \ln b^x + \ln(\ln b)[/tex]
 
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  • #6
Werg22 said:
Is it possible to find a constant value for b in the following equality for any value of x?
[tex] b^{x}(\ln b) = a^x [/tex]
Suppose [tex]a,b\in\mathbb{R}^+[/tex]. If [tex]b^x\ln b=a^x[/tex] then we have
[tex]x\ln b+\ln\ln b=x\ln a[/tex]
or
[tex]\ln\ln b=x(\ln a-\ln b)[/tex].

If [tex]a\neq b[/tex] then this will not hold for all [tex]x\in\mathbb{R}[/tex]. Then we have a=b and [tex]\ln\ln b=0[/tex]. This means the unique solution is [tex]a=b=e[/tex].
 
  • #7
Does that mean that the function a^x has no integral function if a is not equal to e?
 
  • #8
Werg22 said:
Does that mean that the function a^x has no integral function if a is not equal to e?
?
[tex]a^x= e^{ln a^x}= e^{x ln a}[/tex]
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.
 

What is the equation "b^x (ln b) = a^x" used for?

The equation "b^x (ln b) = a^x" is commonly used in mathematics and physics to solve for the value of b when given a specific value for x and a. It is also known as the exponential logarithmic equation.

What is the relationship between b and a in this equation?

The equation "b^x (ln b) = a^x" shows an inverse relationship between b and a. As one variable increases, the other decreases in order to keep the equation balanced.

What is the general method for solving "b^x (ln b) = a^x"?

The general method for solving "b^x (ln b) = a^x" is to use logarithmic properties to isolate b on one side of the equation. This can be done by taking the natural logarithm of both sides and then using algebraic manipulation to solve for b.

Is there a specific range of values for b and a in this equation?

There is no specific range of values for b and a in this equation. However, the equation may become more complex to solve if the values of b and a are not within a reasonable range.

Are there any real-life applications for "b^x (ln b) = a^x"?

Yes, this equation has many real-life applications, such as in calculating the half-life of radioactive substances and in modeling population growth. It is also used in finance to calculate compound interest and in chemistry to determine reaction rates.

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