Substitution of $x=\ln y$

Question

I must be forgetting a log rule, but I'm reading a solution where $$\frac{x^p}{a^x}$$ equals $$\frac{(\ln y)^p}{y^{\ln a}}$$ after substituting $x = \ln y$ but I can't figure out how that works. I can understand $$a^{\ln y }$$ but then how would I obtain the above denominator? To me, the variables just swapped somehow?

Answer 1

As a general rule, amusingly: $$ \ln \left( a^{\ln(b)} \right) = \ln(b) \ln(a) = \ln(a) \ln(b) = \ln \left( b^{\ln(a)} \right) $$ so, raising both sides to the power of $e$, $$ a^{\ln(b)} = b^{\ln(a)} $$ provided that $a,b > 0$.

Answer 2

It sounds like you're asking why $a^{\ln y} = y^{\ln a}$. It's true that it's a formula you can remember, but you can also see it by letting $b = a^{\ln y}$. If I take the logarithm of both sides, I can see that: $$\ln b = \ln (a^{\ln y}) = \ln y \ln a = \ln (y^{\ln a}),$$ and from here you can see that $b = y^{\ln a}$.

Answer 3

Another way to see that for $a, b \gt 0, a^{ \ln b}= b^{\ln a}$, which is completely equivalent to the analyses you've seen in other answers, is to remember that $a=e^{\ln a}$, so for any $t \in \Bbb R, a^t=(e^{\ln a})^t=e^{t \ln a}$. Thus,

$$a^{\ln b}=e^{\ln a \cdot \ln b}= e^{\ln b \cdot \ln a}=b^{\ln a}.$$