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I would like to remove exponentiation in this equation $y=x^t$ and use only multiplication or division. I have $x, t$ and $\ln(x)$. Is it possible?

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$\ln y=t\ln x$ is the best you can do. – David Mitra Feb 6 '13 at 14:56
up vote 1 down vote accepted


This gives you: $\ln(y) = \ln(x^t) = t \ln(x)\quad\implies \quad \ln(y) = t \ln(x)$

Here we use the rule of logarithms (which applies to "$\ln$" as well):

$$\log(a^b) = b \log (a)$$

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So now instead using pow I have to use exp. What if I could also use addition and subtraction? – Martin Feb 6 '13 at 15:25
That's about as simplified as you can get. You could write: $t = \dfrac{\ln(y)}{\ln(x)}$ – amWhy Feb 6 '13 at 15:32
To solve for $y$ given $t$ and $\ln(x)$, evaluate $t\ln(x) = a$, so $\ln(y) = a \implies y = e^a$. I.e. $y = \exp(t\ln(x))$ – amWhy Feb 6 '13 at 15:38

When $a>0$ then we are always allowed to have $$\ln(a^b)=b\ln(a)$$ So take a Neperian logarithm from both sides of your identity.

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What is the Neperian logarithm? Natural log? +1 – amWhy Feb 8 '13 at 0:07
yes. it is the natural one. – Babak S. Feb 8 '13 at 5:02

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