# Exponential expression

• Dec 28th 2013, 11:11 AM
polarbear73
Exponential expression
Can someone give me a hint about transforming $x^{4x}$ into a quotient? I think it involves setting y equal to it and taking a log or something.
• Dec 28th 2013, 11:16 AM
romsek
Re: Exponential expression
can you state the exact problem? What you've written above doesn't make an overwhelming amount of sense.
• Dec 28th 2013, 03:56 PM
HallsofIvy
Re: Exponential expression
Quote:

Originally Posted by polarbear73
Can someone give me a hint about transforming $x^{4x}$ into a quotient? I think it involves setting y equal to it and taking a log or something.

A quotient? What about $\dfrac{1}{x^{-4x}}$?

But I suspect you are really talking about the derivative. If we let $y= x^{4x}$ then $ln(y)= ln(x^{4x})= 4x ln(x)$ so that $\frac{1}{y} y'= 4 ln(x)+ \frac{4x}{x}= 4 ln(x)+ 4$ so that $y'= 4y(ln(x)+ 1)= 4x^4(ln(x)+ 1)$.
• Dec 29th 2013, 08:49 AM
SlipEternal
Re: Exponential expression
Quote:

Originally Posted by HallsofIvy
A quotient? What about $\dfrac{1}{x^{-4x}}$?

But I suspect you are really talking about the derivative. If we let $y= x^{4x}$ then $ln(y)= ln(x^{4x})= 4x ln(x)$ so that $\frac{1}{y} y'= 4 ln(x)+ \frac{4x}{x}= 4 ln(x)+ 4$ so that $y'= 4y(ln(x)+ 1)= 4x^4(ln(x)+ 1)$.

That last equality should be $y' = 4x^{4x}\left(\ln(x) + 1\right)$
• Jan 2nd 2014, 05:22 AM
HallsofIvy
Re: Exponential expression
Yes, thanks.